Here is the chapter's first big surprise, and it is worth pausing on: a wire carrying a current, sitting quietly in a magnetic field, gets pushed. Nobody is touching it. The push comes from the two magnetic fields meeting — the field the current itself makes around the wire, and the field you placed the wire in. On one side of the wire the two fields add up and get stronger; on the other side they fight each other and get weaker. The wire is shoved towards the weaker side, like something squeezed out from between two uneven pressures.
That picture tells you the direction, but for exams you need a rule you can apply in seconds. Fleming's left-hand rule does it: hold your left hand so that your first finger points along the magnetic field (B), your second finger points along the current (I), and your thumb then points along the force (F). Keep the three fingers at right angles to each other, like the corner of a room. There is also a right-hand version in the book — curl the fingers of your right hand from the current towards the field through the smaller angle, and the stretched thumb gives the force — but pick one rule and practise it until it is automatic, because mixing the two hands is the classic exam slip.
Now the size of the force. Experiment says it grows with three things: the current I, the length L of wire actually inside the field, and the strength of the field B. Put together, for a wire at right angles to the field: F = BIL (10.1). Turn that around and you get the definition of field strength itself: B = F / IL. So one tesla is the field that pushes a one-metre wire carrying one ampere, placed at right angles to the field, with a force of one newton. In symbols, 1 T = 1 N A−1 m−1. B is a vector, pointing the same way as the field, and it is also called the magnetic induction or flux density — the same quantity you will meet again in the next section.
If the wire is not at right angles but makes an angle θ with the field, only the part of the wire's length that lies across the field does the pushing, which is L sin θ. So the general rule is F = BIL sin θ (10.2), or in vector form F = IL × B (10.3), where L is a vector along the current. Read the two extremes off that formula: the force is biggest, BIL, when θ = 90°, and it drops to exactly zero when the wire runs along the field, θ = 0°. A wire lying along the field lines feels nothing at all — a fact the exercise will test more than once.
Worked example (book, Example 10.1). A 20.0 cm wire carries 10.0 A in a 0.30 T field at 40° to the field. Convert first: L = 20.0 cm = 0.20 m. Then F = BIL sin θ = 10.0 × 0.30 × 0.20 × sin 40° = 10.0 × 0.30 × 0.20 × 0.643 = 0.39 N. Notice the order of the working: units converted before substituting, angle handled last. Copy that habit and half your numerical slips disappear.
📌 For your information (book, p. 222): magnetism as a subject began with lodestone, a natural mineral found in ancient Türkiye. Lodestone, or magnetite (Fe3O4), attracts iron and steel and lines up with the Earth's magnetic poles — which is how the compass was born. And the chapter opens with the 1820 discovery by Hans Christian Oersted that electricity and magnetism are connected: a current makes a magnetic field around itself, and a changing magnetic field can make a current. That two-way street is this whole chapter.
اس باب کی پہلی بڑی حیرت یہ ہے کہ مقناطیسی میدان میں رکھی ہوئی کرنٹ بردار تار کو دھکا لگتا ہے، حالانکہ اسے کوئی چھو نہیں رہا۔ یہ دھکا دو مقناطیسی میدانوں کے ملنے سے پیدا ہوتا ہے — ایک میدان وہ جو کرنٹ خود تار کے گرد بناتا ہے، اور دوسرا وہ میدان جس میں تار رکھی گئی ہے۔ تار کے ایک طرف دونوں میدان مل کر مضبوط ہو جاتے ہیں، دوسری طرف ایک دوسرے کو کمزور کرتے ہیں، اور تار کمزور طرف دھکیلی جاتی ہے۔
فلیمنگ کا بائیں ہاتھ کا اصول سمت بتاتا ہے: بایاں ہاتھ اس طرح رکھیں کہ پہلی انگلی مقناطیسی میدان (B) کی سمت میں ہو، دوسری انگلی کرنٹ (I) کی سمت میں ہو، تو انگوٹھا قوت (F) کی سمت بتائے گا۔ تینوں انگلیاں ایک دوسرے کے عموداً ہوں۔ قوت کی مقدار کے لیے: تار میدان کے عموداً ہو تو F = BIL، اور عام صورت میں F = BIL sin θ۔ اس سے میدان کی شدت کی تعریف ملتی ہے: B = F / IL، یعنی ایک ٹیسلا وہ میدان ہے جو ایک میٹر لمبی تار میں ایک ایمپیئر کرنٹ پر ایک نیوٹن قوت لگائے۔ جب تار میدان کے ساتھ ساتھ ہو (θ = 0°) تو قوت صفر ہو جاتی ہے۔
مثال: L = 20.0 سینٹی میٹر = 0.20 میٹر، I = 10.0 A، B = 0.30 T، θ = 40° ہو تو F = 10.0 × 0.30 × 0.20 × sin 40° = 0.39 N۔ پہلے اکائیاں بدلیں، پھر قیمتیں رکھیں — یہ ترتیب غلطیوں سے بچاتی ہے۔
If field lines are a picture of a magnetic field, then magnetic flux is a way of counting that picture. Imagine field lines as threads passing through a window. The flux through the window is simply how many of those threads pass through it. More lines through the same window means more flux; tilt the window so the lines slide past it instead of through it, and the flux falls.
Formally, if B is the number of lines per unit area placed at right angles to the field, then the total flux through an area A that is perpendicular to the field is ΦB = BA, or rearranged, B = ΦB / A (10.4). That is why B is called the flux density: it is the flux per unit area. Real surfaces are rarely square-on to the field, though. A surface has an area vector A whose size is the area and whose direction is the normal (the perpendicular) to the surface. If that normal makes an angle θ with B, only the component of B along the normal, B cos θ, actually threads the surface. So the general rule is ΦB = BA cos θ (10.5), or as a dot product, ΦB = B · A (10.6). Flux is a scalar — it has size but no direction of its own — even though it is built from two vectors.
Two orientations are worth memorising because they come up everywhere. When the field runs along the normal to the surface (θ = 0°, cos θ = 1), the flux is at its maximum, BA. When the field lies in the plane of the surface (θ = 90°, cos θ = 0), no lines thread it at all and the flux is zero. A loop lying flat in a field can therefore have zero flux even while sitting in a strong field — and a loop that is turned through a field changes its flux continuously, which is exactly what the rest of the chapter exploits. For a curved surface in a uniform field, the book's trick is to chop it into tiny flat pieces, work out the flux through each, and add them up.
Units follow from the definition of the tesla: flux has the unit weber (Wb), where 1 Wb = 1 T m2 = 1 N m A−1. And flux density, being flux per unit area, is measured in Wb m−2, which is just the tesla again (1 T = 1 Wb m−2 = 1 N A−1 m−1).
Worked example (book, Example 10.2). A rectangular loop 25 cm by 20 cm sits in a 1.2 T field. Area first: A = 0.25 × 0.20 = 5 × 10−2 m2. The three orientations in the book's figure give the angle between B and the area vector as 60°, 45° and 30° (note carefully: these are the angles to the normal, not to the loop itself — the loop's own tilt in the drawing is the complement). Then: (a) Φ = 1.2 × 5 × 10−2 × cos 60° = 1.2 × 5 × 10−2 × 0.5 = 3.0 × 10−2 Wb; (b) Φ = 1.2 × 5 × 10−2 × 0.707 = 4.2 × 10−2 Wb; (c) Φ = 1.2 × 5 × 10−2 × 0.866 = 5.2 × 10−2 Wb. The smaller the angle to the normal, the more flux — exactly as the counting-lines picture suggests.
📌 Interesting information (book, p. 224): the Earth's own magnetic field is approximately that of a giant bar magnet, or dipole, buried inside the planet — with the Earth's south magnetic pole towards the geographic north pole, and its north magnetic pole towards the geographic south pole. That is why the north-seeking end of your compass points north: it is being attracted to a magnetic south pole sitting under the Arctic.
مقناطیسی میدان کی لکیروں کو دھاگوں کی طرح سمجھیں جو کسی کھڑکی سے گزرتے ہیں۔ مقناطیسی فلکس یہ گنتی ہے کہ کتنی لکیریں اس رقبے سے گزرتی ہیں۔ اگر رقبہ میدان کے عموداً ہو تو Φ = BA، اور عام صورت میں رقبے کے نارمل اور میدان کے درمیان زاویہ θ ہو تو Φ = BA cos θ، یعنی Φ = B · A۔ فلکس ایک اسکیلر مقدار ہے۔
دو حالتیں یاد رکھیں: جب میدان نارمل کے ساتھ ہو (θ = 0°) تو فلکس زیادہ سے زیادہ یعنی BA ہوتا ہے، اور جب میدان رقبے کی سطح میں ہو (θ = 90°) تو فلکس صفر ہوتا ہے۔ اسی لیے ایک لوپ مضبوط میدان میں بھی صفر فلکس رکھ سکتا ہے۔ فلکس کی اکائی ویبر (Wb) ہے، اور فلکس کثافت B کی اکائی Wb m−2 یعنی ٹیسلا ہے۔
مثال: لوپ 25 سینٹی میٹر × 20 سینٹی میٹر = 5 × 10−2 میٹر2، B = 1.2 T۔ نارمل سے زاویے 60°، 45° اور 30° پر فلکس بالترتیب 3.0 × 10−2 Wb، 4.2 × 10−2 Wb اور 5.2 × 10−2 Wb بنتا ہے۔ خیال رہے کہ یہ زاویے نارمل سے ہیں، لوپ کے اپنے جھکاؤ سے نہیں۔
One loop threading some flux is a small effect. Wind that loop a hundred times and the effect multiplies — and that multiplication is the whole idea of flux linkage. If a coil has N turns and each turn threads the same flux ΦB, then the flux linkage of the coil is Φ = NΦB (10.7). Think of it as the total amount of flux the coil as a whole "holds on to": every turn catches the same lines, so the coil catches them N times over.
This is not just bookkeeping. Flux linkage is the quantity that actually drives electromagnetic induction, which is why it shows up in transformers, motors, generators and inductors — every device where a coil and a changing magnetic field do useful work. When the flux linkage changes, an emf appears (that is Faraday's law, coming up in section 10.6), and the bigger N is, the bigger the emf. That is the practical reason transformers and generator coils are wound with many turns instead of one fat loop: turns are cheap, and each one adds its share to the linkage.
📌 For your information (book, p. 227): the magnetic strip on an ATM card works on the same idea in miniature — it holds millions of tiny magnetic domains glued together in a resin binder, and the card machine reads the information encoded on that strip to give you access to your account. Small magnets, carefully arranged, carrying real information.
ایک لوپ سے گزرنے والا فلکس چھوٹا اثر ہوتا ہے، لیکن اگر اسی لوپ کو سو بار لپیٹ دیا جائے تو اثر کئی گنا بڑھ جاتا ہے۔ یہی فلکس لنکیج ہے: اگر کوائل میں N چکر ہوں اور ہر چکر سے ΦB فلکس گزرے تو کل فلکس لنکیج Φ = NΦB ہوگی۔ ہر چکر وہی لکیریں پکڑتا ہے، اس لیے کوائل مجموعی طور پر انہیں N بار پکڑتی ہے۔
یہی مقدار برقی مقناطیسی انڈکشن کو چلاتی ہے، اس لیے ٹرانسفارمر، موٹر، جنریٹر اور انڈکٹر میں یہی تصور کام آتا ہے۔ چکر زیادہ ہوں تو انڈیوسڈ ای ایم ایف بھی زیادہ ہوتی ہے — یہی وجہ ہے کہ ان آلات میں ایک موٹے لوپ کے بجائے کئی چکر والی کوائل استعمال ہوتی ہے۔
A current is nothing but charges on the move, so the force on a wire must really be the sum of the forces on the individual charges inside it. The book proves it neatly: a conductor of length L carrying N charges of charge q in time t carries current I = Nq / t, and with the charges drifting at speed v, that becomes I = Nqv / L. Feed that into F = ILB and the N and L cancel down to the force on one particle: F = q(v × B) (10.13). Its size is F = qvB sin θ (10.14), where θ is the angle between the velocity and the field. Maximum when the particle crosses the field at right angles (θ = 90°), zero when it travels along the field (θ = 0°). The direction comes from Fleming's left-hand rule or the right-hand rule for the cross product — and for a negative charge, the force flips to the opposite side.
That one fact explains the two pictures in the book. A positive particle entering a field that points into the page is pushed upwards and curves away in a rising arc; a negative particle entering the same field is pushed downwards and curves the other way. And here is the beautiful consequence: because the force is always at right angles to the velocity, it can bend the path but it can never speed the particle up or slow it down. It does no work. So a particle crossing a uniform field at right angles goes round in a circle, with the magnetic force supplying the centripetal force: qvB = mv2 / r, giving radius r = mv / (qB). Faster particles swing wider, stronger fields curl them tighter, and heavier particles need a bigger circle than light ones at the same speed — a sentence that answers several exercise questions by itself.
Worked example (book, Example 10.3). An electron enters a uniform field perpendicularly at 104 m s−1 (B = 2.5 Wb m−2, m = 9.11 × 10−31 kg, e = 1.6 × 10−19 C). Since the velocity is perpendicular to the field, the force is perpendicular to the velocity and acts as centripetal force: evB = mv2 / r, so r = mv / (eB) = (9.11 × 10−31 × 104) / (1.6 × 10−19 × 2.5) = 9.11 × 10−27 / 4.0 × 10−19 = 2.3 × 10−8 m. The electron's path is a circle of that radius — tiny, because the electron is so light.
📌 Do you know? (book, p. 227): like electric field lines, magnetic field lines never cross each other — though they can attract or repel each other. 📌 Point to ponder! (book, p. 228): why does a picture become distorted when a magnet is brought near an old TV, computer monitor or oscilloscope screen? Because those screens paint the picture with a beam of electrons, and the magnet's field deflects the beam — the same qvB force, bending the electrons away from where they were meant to land. (This answer is worked out from the chapter's force law; the book poses it as a question, it does not print the answer.)
کرنٹ دراصل حرکت کرتے ہوئے چارجز ہی ہیں، اس لیے تار پر لگنے والی قوت اندر کے ہر چارج پر لگنے والی قوتوں کا مجموعہ ہے۔ ایک ذرے پر قوت F = q(v × B) ہے، اور اس کی مقدار F = qvB sin θ ہے۔ جب ذرہ میدان کے عموداً حرکت کرے تو قوت زیادہ سے زیادہ ہوتی ہے، اور جب میدان کے ساتھ ساتھ چلے تو قوت صفر ہوتی ہے۔ منفی چارج پر قوت مخالف سمت میں لگتی ہے۔
قوت ہمیشہ رفتار کے عموداً لگتی ہے، اس لیے یہ ذرے کی رفتار نہیں بدل سکتی، صرف راستہ موڑ سکتی ہے۔ یہی وجہ ہے کہ یکساں میدان میں عموداً داخل ہونے والا ذرہ دائرے میں گھومتا ہے، اور اس دائرے کا رداس r = mv / qB ہوتا ہے۔ تیز ذرہ بڑا دائرہ بناتا ہے، مضبوط میدان دائرہ چھوٹا کرتا ہے۔
مثال: الیکٹران v = 104 m s−1، B = 2.5 T ہو تو r = (9.11 × 10−31 × 104) / (1.6 × 10−19 × 2.5) = 2.3 × 10−8 m۔ الیکٹران اتنا ہلکا ہے اس لیے دائرہ بہت چھوٹا بنتا ہے۔
A velocity selector is a filter for speed, and it works by setting two forces against each other until only one speed survives. Picture a tube sitting in a magnetic field, with a pair of parallel plates inside it making a uniform electric field at right angles to the magnetic one. A positive charge flying through feels two pushes: the magnetic force qvB in one direction, and the electric force qE in the opposite direction. For most speeds one push wins and the particle is deflected off course. But for exactly one speed the two pushes cancel — qvB = qE, the q cancels, and the selected speed is simply v = E / B (10.15). That particle sails straight through and out the far end, untouched. Every other speed is bent away and never reaches the exit. Notice something elegant: the selected speed does not depend on the particle's charge or mass at all — only on the two fields you set.
Worked example (book, Example 10.4). Alpha particles with speeds from 1000 to 2000 m s−1 enter a selector with E = 300 V m−1 (= 300 N C−1) and B = 0.20 T. The undeflected speed is v = E / B = 300 / 0.20 = 1500 m s−1. Only the alpha particles travelling at 1500 m s−1 get through; slower ones are pulled one way by the electric field, faster ones the other way by the magnetic field.
Worked example (book, Example 10.5). A particle crosses a selector undeflected with E = 4.8 × 103 N C−1 and B = 0.2 T. First its speed: v = E / B = 4.8 × 103 / 0.2 = 2.4 × 104 m s−1. Then the electric field is switched off, and the particle circles in the magnetic field alone with radius r = 3.0 cm = 3.0 × 10−2 m. The magnetic force now supplies the centripetal force, Bqv = mv2 / r, so q / m = v / (Br) = E / (B2r) = 4.8 × 103 / (0.04 × 3.0 × 10−2) = 4.8 × 103 / 1.2 × 10−3 = 4 × 106 C kg−1. This two-step trick — selector first to find v, then a magnetic circle to find q/m — is a favourite exam construction, so learn the pair together.
📌 Point to ponder! (book, p. 230): a force is exerted on a moving charged particle in a magnetic field — in what direction should it move if the force is not exerted on it? Answer: along the field itself (or exactly opposite to it). Then θ = 0° or 180°, sin θ = 0, and qvB sin θ = 0. Standing still also gives no force, since v = 0, but the question's point is the direction: parallel to the field lines, the particle slips through untouched. (Worked out from F = qvB sin θ; the book poses it as a question.)
ویلاسٹی سلیکٹر رفتار کا فلٹر ہے: اس میں برقی میدان اور مقناطیسی میدان ایک دوسرے کے عموداً لگائے جاتے ہیں۔ چارج ذرے پر دو قوتیں لگتی ہیں — مقناطیسی قوت qvB ایک طرف، برقی قوت qE دوسری طرف۔ صرف ایک رفتار پر دونوں قوتیں برابر ہوتی ہیں: qvB = qE، یعنی v = E / B۔ وہ ذرہ سیدھا گزر جاتا ہے، باقی سب مڑ جاتے ہیں۔ خوبصورت بات یہ ہے کہ یہ رفتار چارج یا کمیت پر منحصر نہیں، صرف دونوں میدانوں پر ہے۔
مثال: E = 300 N C−1، B = 0.20 T ہو تو منتخب رفتار v = 300 / 0.20 = 1500 m s−1۔ دوسری مثال: پہلے v = E / B = 4.8 × 103 / 0.2 = 2.4 × 104 m s−1، پھر دائرے سے q/m = v / Br = 4 × 106 C kg−1۔ یہ دو مرحلوں والی ترکیب امتحان میں بہت آتی ہے۔
So far the field has pushed things that were already moving. Now the story flips: move a conductor across a magnetic field, and a voltage appears between its ends — an induced emf, acting just like a battery. Connect the ends into a loop and a current flows, with no battery anywhere in the circuit. The emf made by motion alone is called motional emf.
The book's experiment makes the mechanism visible. A rod of length L slides along two rails in a field pointing into the page, with a galvanometer completing the loop. As the rod moves right at speed v, each charge inside it moves with it and feels qvB, pushing charges to one end of the rod. Charges pile up until their own electric field balances the magnetic push: qE = qvB, so E = vB. The voltage between the ends is then ε = ΔV = LvB (10.19) — or, if the motion is at an angle θ to the field, ε = vBL sin θ (10.20). Once current flows, the pile-up drains, the balance breaks, and the magnetic push keeps topping the charge up at the top end — so the current keeps flowing as long as the rod keeps moving. Stop the rod, and the whole thing dies.
Faraday's law generalises this one experiment into the chapter's master rule. As the rod moves a distance Δx in time Δt, the loop's area grows by ΔA = LΔx, so the flux through it grows by ΔΦB = BΔA. Substituting v = Δx / Δt into ε = vBL gives ε = −ΔΦB / Δt (10.23) for one loop, and for a coil of N loops, ε = −N ΔΦB / Δt (10.24). In words: the average emf induced in a coil of N loops equals the negative of the rate at which the magnetic flux through the coil is changing. The minus sign is not decoration — it carries the direction, and the next section explains why it matters. Note that what counts is the rate of change of flux, not the flux itself: a big steady flux induces nothing, while a small flux changing fast induces a lot.
Worked example (book, Example 10.6). A 25 cm rod moves at 0.5 m s−1 at right angles to a 0.25 T field. Convert: L = 0.25 m. Then ε = vBL = 0.5 × 0.25 × 0.25 = 3.13 × 10−2 V. Worked example (book, Example 10.7). A loop in a 0.6 T field perpendicular to its plane shrinks at a constant rate ΔA / Δt = 0.8 m2 s−1. The rate of change of flux is B × ΔA / Δt = 0.6 × 0.8 = 0.48 Wb s−1, so with N = 1, the magnitude of the induced emf is 0.48 V. In both examples, check the units as you go: tesla times metres times metres per second gives volts, which is a quick way to catch a slipped decimal.
📌 Do you know? (book, p. 232): wireless charging works on electromagnetic induction — the charger and the phone each hide a coil, and the changing flux between them transfers the energy without any wire. 📌 Point to ponder! (book, p. 232): an induction heater boils water in a metal pot but leaves a glass pot cold, even though the heater's top feels cool to the touch. Why? The coil under the top carries alternating current, producing a changing magnetic field; that changing flux induces currents in the metal pot, and those currents heat it. Glass is an insulator, so no current is induced in it and it stays cold. (Worked out from Faraday's law; the book poses it as a question.)
اب کہانی الٹ جاتی ہے: کسی موصل کو مقناطیسی میدان میں حرکت دیں تو اس کے سروں کے درمیان وولٹیج پیدا ہو جاتی ہے — یہ انڈیوسڈ ای ایم ایف ہے، جو بیٹری کی طرح کام کرتی ہے۔ صرف حرکت سے پیدا ہونے والی ای ایم ایف کو موشنل ای ایم ایف کہتے ہیں۔
راڈ کے اندر ہر چارج حرکت کے ساتھ qvB قوت محسوس کرتا ہے، چارج ایک سرے پر جمع ہوتے ہیں، یہاں تک کہ ان کا اپنا برقی میدان مقناطیسی دھکے کو برابر کر دے: qE = qvB، یعنی E = vB، اور سروں کے درمیان وولٹیج ε = LvB بنتی ہے۔ فیراڈے کا قانون اسے عام شکل دیتا ہے: ε = −N ΔΦ / Δt، یعنی کوائل میں اوسط انڈیوسڈ ای ایم ایف فلکس کی تبدیلی کی شرح کے منفی کے برابر ہے۔ اہم بات یہ ہے کہ فلکس کی مقدار نہیں، اس کی تبدیلی کی شرح ای ایم ایف بناتی ہے — بڑا مگر ساکن فلکس کچھ پیدا نہیں کرتا۔
مثال: L = 0.25 m، v = 0.5 m s−1، B = 0.25 T ہو تو ε = 0.5 × 0.25 × 0.25 = 3.13 × 10−2 V۔ دوسری مثال: B = 0.6 T اور رقبے کی تبدیلی 0.8 m2 s−1 ہو تو ε = 0.6 × 0.8 = 0.48 V۔
Faraday's law tells you how big the induced emf is. It takes a second law to tell you which way the current goes, and that law has a beautifully stubborn character: the direction of the induced current is always such that it opposes the change that causes the current. That is Lenz's law, from the Russian physicist Heinrich Lenz in 1834. Note the careful wording, because it is where students go wrong: the induced current opposes the change in flux, not the flux itself. If flux is increasing, the induced current makes a field that tries to reduce it; if flux is decreasing, the induced current makes a field that tries to top it back up. Opposing the change, in both cases.
The book's magnet-and-coil experiment shows it cleanly. Push a bar magnet's north pole towards a coil, and the coil's near face becomes a north pole too — the two north poles repel, fighting the push. By the right-hand rule, that means the induced current flows anticlockwise as seen from the magnet. Pull the magnet away, and the coil flips: its near face becomes a south pole, attracting the retreating magnet and fighting the pull. The current acts against whatever you are doing, push or pull. Lenz's law applies directly to induced currents in closed loops; for an open loop you imagine it closed, find the current's direction, and read off the emf's direction from that.
There is a deeper reason the law must hold, and the book proves it with the sliding rod. When the rod moves right, the induced current runs anticlockwise round the loop, and the magnetic force on the current-carrying rod, Fm = ILB sin 90°, points against the rod's motion. To keep the rod moving at constant speed you must push it with an equal external force — and that push is where the induced current's energy comes from. Your hand does mechanical work, and it reappears as electrical energy in the circuit. Now imagine Lenz's law were reversed: the induced current would push the rod forwards, speeding it up, inducing a bigger current, pushing harder still — motion feeding on itself, energy appearing from nowhere. That self-perpetuating machine is impossible, so Lenz's law is really the law of conservation of energy wearing a different coat. If you ever forget the direction rule, that energy argument will rebuild it for you.
📌 Point to ponder! (book, p. 233): a copper ring falls through a rectangular region of constant magnetic field directed into the page. What can you say about the current in the ring at positions 2, 3 and 4 of the book's figure? Worked out from Faraday's law (the book poses it as a question): at position 2, where the ring is entering the field, its flux is increasing, so a current is induced to oppose the increase. At positions 3 and 4, where the ring is fully inside the uniform field, its flux is constant, so no current flows at all — a steady flux, however large, induces nothing. When it leaves the field below, a current appears again, in the opposite sense, to oppose the decrease. 📌 Point to ponder! (book, p. 235): by neglecting the resistance, can a constant current in a coil set up a potential difference across the coil? No: a constant current means a constant flux, and a constant flux induces no emf. With resistance neglected there is no resistive voltage drop either, so no potential difference appears across the coil. (Both answers are derived from the chapter's laws, not printed answers in the book.)
📌 For your information (book, p. 233): Faraday's homopolar generator — a disc spinning in a magnetic field — was the machine with which he first produced a continuous induced current, the ancestor of every generator since.
فیراڈے کا قانون ای ایم ایف کی مقدار بتاتا ہے، سمت بتانے کے لیے لینز کا قانون ہے: انڈیوسڈ کرنٹ کی سمت ہمیشہ ایسی ہوتی ہے کہ وہ اس تبدیلی کی مخالفت کرے جس سے کرنٹ پیدا ہوا ہو۔ غور کریں: مخالفت فلکس کی نہیں، فلکس کی تبدیلی کی ہوتی ہے۔ فلکس بڑھ رہا ہو تو انڈیوسڈ کرنٹ اسے کم کرنے کی کوشش کرتا ہے، اور فلکس کم ہو رہا ہو تو اسے بڑھانے کی کوشش کرتا ہے۔
مقناطیس کو کوائل کی طرف دھکیلیں تو کوائل کا قریبی سرا بھی شمالی قطب بن جاتا ہے اور دھکے کی مخالفت کرتا ہے؛ مقناطیس کو کھینچیں تو وہ سرا جنوبی قطب بن کر کھنچاؤ کی مخالفت کرتا ہے۔ یہی قانون توانائی کے تحفظ کا قانون ہے: سلائیڈنگ راڈ کی مثال میں انڈیوسڈ کرنٹ راڈ کی حرکت کی مخالف سمت میں قوت لگاتا ہے، اس لیے راڈ کو چلائے رکھنے کے لیے باہر سے کام کرنا پڑتا ہے — اور وہی کام برقی توانائی بن جاتا ہے۔ اگر لینز کا قانون الٹ ہوتا تو حرکت خود بخود تیز ہوتی چلی جاتی اور توانائی کہیں سے پیدا ہو جاتی، جو ناممکن ہے۔
Faraday's law, ε = −N ΔΦ / Δt, already contains the whole list of what makes an induced emf bigger or smaller — you just have to read it as a recipe. The book names three factors, and each one falls straight out of the formula.
1. Rate of change of magnetic flux. Faster changes give bigger emfs. Shove the magnet in quickly, and the galvanometer kicks hard; slide it in slowly, and it barely trembles. It is the speed of the flux change that counts, not the size of the flux — a coil sitting in a huge steady field has no emf at all.
2. Number of turns of the coil. The emf is proportional to N. Double the turns and you double the emf, because each turn contributes its own share of the changing linkage. This is the cheapest way to get a bigger emf in a generator or transformer: wind more turns.
3. Relative speed. How fast the coil (or conductor) moves through the field sets how fast the flux changes. Faster motion means a faster flux change, and therefore a larger emf. In the motional case this is visible directly in ε = vBL: double the speed, double the emf.
Keep the three together as one sentence for revision: big emf needs fast flux change, many turns, and fast relative motion. Every generator design question in the exercise is, at bottom, asking which of these three you would change.
فیراڈے کے قانون ε = −N ΔΦ / Δt میں ہی ای ایم ایف کو بڑھانے یا گھٹانے والے تینوں عوامل چھپے ہیں۔ ۱۔ فلکس کی تبدیلی کی شرح: تبدیلی جتنی تیز ہوگی، ای ایم ایف اتنی ہی زیادہ ہوگی — بڑا مگر ساکن فلکس کوئی ای ایم ایف پیدا نہیں کرتا۔ ۲۔ کوائل کے چکروں کی تعداد: ای ایم ایف N کے متناسب ہے، چکر دوگنے کریں تو ای ایم ایف دوگنی ہو جاتی ہے۔ ۳۔ نسبتی رفتار: کوائل یا موصل جتنی تیزی سے میدان میں حرکت کرے گا، فلکس اتنی تیزی سے بدلے گا اور ای ایم ایف اتنی ہی زیادہ ہوگی؛ موشنل صورت میں ε = vBL سے یہ براہ راست نظر آتا ہے۔
یاد رکھنے کا ایک جملہ: بڑی ای ایم ایف کے لیے تیز فلکس تبدیلی، زیادہ چکر اور تیز نسبتی حرکت چاہیے۔
Some materials refuse to pick a side, and ferrofluid is one of them: a liquid that behaves like a magnet. It is a colloidal suspension of tiny magnetic particles in a carrier fluid such as oil or water. The particles are usually iron oxide, ground down to about 10 nanometres across, and each one is coated with a surfactant — a substance that lowers surface tension and stops the particles from clumping together. Their tiny size, the constant jiggling of Brownian-style motion, and the thickness of the fluid all stop them from settling out, so the mixture stays evenly dispersed instead of separating like sand in water.
Bring a magnet near, and the trick reveals itself. With no magnet around, the fluid acts like an ordinary liquid. With a magnet nearby, the particles are temporarily magnetised and the whole fluid becomes a magnet: the particles line up in chains along the field lines, and the competition between magnetic pull, surface tension and gravity sculpts the surface into the famous spikes. Where the magnetic force beats the other two, the fluid stands up in stiff, solid-looking points; take the magnet away and the particles demagnetise, the spikes collapse, and it is a liquid again. It never stays magnetic on its own — it borrows its magnetism for exactly as long as the field is there.
The book's experiment. You need laser-printer toner (about 40% iron oxide in nanometre-sized particles), cooking oil, a test tube, a glass bottle, a small stick and a magnet. Stir some toner into the oil to make the ferrofluid, pour it into the bottle, and shake it: it behaves like a liquid. Now bring the magnet near the outside of the bottle, and the fluid jumps towards the magnet — it has itself become a magnet. Hold the magnet against the side of the bottle and you will see the spiky structure form, just as in the book's photograph.
Applications. In electronics, ferrofluids seal the rotating shafts of computer hard drives and other rotating motors, and in loudspeakers they sit around the voice coil, carrying heat away and damping vibrations to improve the sound. In medicine, an external magnet can steer ferrofluid-carried drugs to a tumour or another target site, cutting side effects, and ferrofluids can serve as contrast agents in MRI scans. Engineers also use them to damp or precisely control the flow of liquids by shaping the magnetic field. A fluid you can push around with a magnet turns out to be useful in a surprising number of places.
📌 For your information (book, p. 237): the first ferrofluid was developed by NASA in 1960, from natural magnetite. It was invented to move liquids through space, where there is no gravity to make fluids flow downhill — a magnet could do the pushing instead.
فیرو فلوئڈ ایک ایسا مائع ہے جو مقناطیس کی طرح برتاؤ کرتا ہے۔ یہ دراصل کسی مائع (جیسے تیل یا پانی) میں باریک مقناطیسی ذرات کی کولائیڈل معلقیت ہے۔ یہ ذرات عام طور پر آئرن آکسائیڈ کے ہوتے ہیں، تقریباً 10 نینومیٹر جسامت کے، اور ہر ذرے پر سرفیکٹنٹ کی تہہ ہوتی ہے جو انہیں آپس میں جمنے سے روکتی ہے۔ ذرات کی چھوٹی جسامت اور مسلسل حرکت انہیں نیچے بیٹھنے نہیں دیتی، اس لیے آمیزہ یکساں رہتا ہے۔
مقناطیس قریب لائیں تو ذرات عارضی طور پر مقناطیسی ہو جاتے ہیں اور پورا مائع مقناطیس بن جاتا ہے: ذرات میدان کی لکیروں کے ساتھ زنجیروں میں لگ جاتے ہیں، اور مقناطیسی قوت، سطحی تناؤ اور کشش ثقل کی کشمکش سے سطح پر مشہور سپائکس بن جاتے ہیں۔ مقناطیس ہٹائیں تو ذرات دوبارہ غیر مقناطیسی ہو جاتے ہیں اور مائع پھر عام مائع بن جاتا ہے — یہ خود کبھی مستقل مقناطیس نہیں رہتا۔
کتاب کا تجربہ: لیزر پرنٹر ٹونر (جس میں تقریباً 40 فیصد نینومیٹر جسامت کا آئرن آکسائیڈ ہوتا ہے) کو کھانے کے تیل میں ملا کر فیرو فلوئڈ بنائیں، بوتل میں ڈالیں اور ہلائیں تو یہ عام مائع لگے گا؛ مقناطیس قریب لائیں تو مائع مقناطیس کی طرف لپکے گا اور بوتل کی دیوار پر سپائکس بن جائیں گے۔ استعمالات: ہارڈ ڈرائیو اور موٹروں کی سیل، لاؤڈ اسپیکر میں وائس کوائل کی ٹھنڈک اور آواز کی بہتری، دواؤں کو مقناطیس سے ٹیومر تک پہنچانا، ایم آر آئی میں کنٹراسٹ، اور مائعات کے بہاؤ کا کنٹرول۔
An earthquake is a shaking of the ground, and a seismometer is the instrument that listens to it — by turning ground motion into an electrical signal, using exactly the induction of this chapter. A typical seismometer has a heavy weight suspended by a spring from a frame that is fixed to the ground. When an earthquake comes, the ground — and the frame with it — moves, but the weight, by its inertia, tends to stay where it is. That gives relative motion between the weight and the frame. The weight is usually attached to a magnet, which therefore moves inside a coil of wire fixed to the frame. By Faraday's law, that changing flux induces an emf in the coil, and an induced current flows.
The learning objectives put the key point precisely: the emf induced in the coil is directly proportional to the displacement associated with the ground motion — or, in the book's running text, the induced current is proportional to the velocity of the ground motion. Either way, the electrical signal is a faithful copy of the shaking, and it is amplified and recorded to give data on the earthquake's amplitude, frequency and duration, from which its location, magnitude and depth are worked out. In practice the instrument is buried underground at a depth of 50–1000 metres, inside a protective case called a vault — a cylindrical steel tank about 1 metre wide and 2 metres deep, set on a concrete pad — so that surface noise does not swamp the signal it is trying to hear.
📌 For your information (book, p. 238): most earthquakes are caused by plate tectonics (displacement of plates) and occur at a depth of up to 60 km; they are categorised as shallow, intermediate (as deep as 280 km beneath the crust) and deep (past 280 km). 📌 For your information (book, p. 239): there are two main types of seismic waves: P-waves, which are primary and longitudinal, and S-waves, which are secondary and transverse.
سیسمومیٹر زلزلے کی حرکت کو سننے والا آلہ ہے، اور یہ اسی باب کی انڈکشن سے کام کرتا ہے۔ اس میں ایک بھاری وزن اسپرنگ سے لٹکا ہوتا ہے، جو زمین سے جڑے فریم سے معلق ہوتا ہے۔ زلزلہ آئے تو زمین اور فریم حرکت کرتے ہیں، لیکن وزن اپنی جمود (inertia) کی وجہ سے اپنی جگہ رہنے کی کوشش کرتا ہے — یوں وزن اور فریم کے درمیان نسبتی حرکت پیدا ہوتی ہے۔ وزن عام طور پر ایک مقناطیس سے جڑا ہوتا ہے جو فریم سے جڑی کوائل کے اندر حرکت کرتا ہے، اور فیراڈے کے قانون سے کوائل میں ای ایم ایف پیدا ہو جاتی ہے۔
یہ برقی سگنل زمین کی حرکت کی وفادار نقل ہوتا ہے: کوائل میں پیدا ہونے والی ای ایم ایف زمین کی حرکت سے وابستہ ڈسپلیسمنٹ کے راست متناسب ہوتی ہے، اور اسے بڑھا کر ریکارڈ کیا جاتا ہے تاکہ زلزلے کی شدت، تعدد، دورانیہ، مقام اور گہرائی معلوم ہو سکے۔ عملی طور پر یہ آلہ 50 سے 1000 میٹر گہرائی میں زمین کے اندر، والٹ نامی حفاظتی اسٹیل ٹینک (تقریباً 1 میٹر چوڑا، 2 میٹر گہرا) میں رکھا جاتا ہے تاکہ سطح کا شور سگنل کو خراب نہ کرے۔
The book's exercise (printed pages 239–243) has five parts: Multiple Choice Questions (10), Short Answer Questions (9), Constructed Response Questions (9), Comprehensive Questions (6) and Numerical Problems (12) — 46 questions in all, every one answered below. There is no separate inquisitive or review section in this chapter's exercise, so none is invented here. Where a question goes beyond a single book passage, the answer says so plainly instead of pretending the book said it. ⚠️ Three points need care, and each is flagged at its own question: in MCQ 10.1 the words "over the wire" are read as "directly above the wire"; in MCQ 10.7 option (a) is the general formula, but the value asked for is zero; and in Numerical 10.9 the book's figure is small, so the rod identification follows the book's printed answer key.
10.1. A current is flowing towards north along a power line. The direction of the magnetic field over the wire is directed towards:
Answer: (c) east — Reading "over the wire" as directly above it: the field circles the wire, and at a point above a northward current the tangent to those circles points east (from Biot–Savart, the current direction crossed with the upward direction gives east). If "over" were read differently the answer would change, so the reading is stated here rather than hidden. تار کے عین اوپر، شمال کی طرف بہتے کرنٹ کا مقناطیسی میدان مشرق کی طرف ہوتا ہے۔
10.2. The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to:
Answer: (b) the particle's momentum — From r = mv / qB, the radius grows with mv (momentum) and shrinks with q and B. Energy is not in the formula directly, and charge sits in the denominator. r = mv / qB سے رداس مومینٹم کے راست متناسب ہے۔
10.3. The diagram shows a beam of electrons entering a magnetic field. What is the effect of magnetic field on the electrons?
Answer: (b) They are deflected out of the plane of the diagram. — In the book's figure the beam travels rightwards and the field points downwards in the plane. For a positive charge, v × B would point into the page; electrons carry negative charge, so the force flips to out of the page. The sign of the charge is the whole question. الیکٹران منفی چارج رکھتا ہے، اس لیے قوت صفحے سے باہر کی طرف لگتی ہے۔
10.4. The force exerted on a wire of 1 metre length carrying 1 ampere current placed at right angle to the magnetic field is called:
Answer: (c) magnetic induction — This is the book's own definition of B (section 10.1): B = F / IL, so a 1 m wire with 1 A at right angles feels a force numerically equal to B, and B is the magnetic induction (flux density). یہ B یعنی مقناطیسی انڈکشن کی تعریف ہے۔
10.5. The unit of flux density is:
Answer: (a) NA−1m−1 — That is the tesla in base units: 1 T = 1 N A−1 m−1 = 1 Wb m−2. The other options fail a unit check against B = F / IL. ٹیسلا = N A−1 m−1، یعنی Wb m−2۔
10.6. A moving charged particle is surrounded by:
Answer: (c) both electric and magnetic field — Its charge alone makes an electric field, and its motion adds a magnetic field around it. A stationary charge has only the electric field, which is why "moving" is the key word. چارج سے برقی میدان اور حرکت سے مقناطیسی میدان — دونوں موجود ہوتے ہیں۔
10.7. Magnetic force on the charge q moving parallel to magnetic field with velocity v is:
Answer: (c) zero — ⚠️ Option (a) is the general formula F = qvB sin θ, but for parallel motion θ = 0° and sin θ = 0, so the value is zero. The question asks for the force in this case, and that value is (c). متوازی حرکت میں θ = 0° ہوتا ہے، اس لیے F = qvB sin 0° = 0۔
10.8. The unit NA−1m−1 is called:
Answer: (b) tesla — NA−1m−1 is the tesla, the unit of B. The weber is the unit of flux (Wb = T m2), and the coulomb is the unit of charge. NA−1m−1 ٹیسلا کہلاتا ہے؛ ویبر فلکس کی اکائی ہے۔
10.9. Electrons while moving perpendicularly through a uniform magnetic field are:
Answer: (c) deflected along circular path — Perpendicular entry makes the force always perpendicular to the velocity, so it acts as a centripetal force and the electron circles (section 10.4). "North pole" and "south pole" do not describe the direction of a circular deflection at all. عموداً داخل ہونے پر قوت سینٹری پیٹل قوت بن جاتی ہے اور الیکٹران دائرے میں گھومتا ہے۔
10.10. A magnet is suspended from a spring, and moves in and out of the coil connected to a galvanometer. When the magnet oscillates, the galvanometer shows:
Answer: (d) deflection to the left and right, but the amplitude steadily decreases — Each swing changes the flux through the coil, first one way then the other, so the induced current alternates left and right. But the oscillation itself loses energy (air resistance, and the induced current's own opposing force), so the swings — and with them the deflections — shrink steadily. Constant amplitude would need a lossless, perpetually driven system. ہر جھولے پر فلکس ایک بار بڑھتا اور ایک بار گھٹتا ہے، اس لیے سوئی دونوں طرف جھولتی ہے، مگر توانائی ضائع ہونے سے جھولے آہستہ آہستہ چھوٹے ہوتے جاتے ہیں۔
Lenz's law is a statement about the direction of a current: the induced current flows so as to oppose the flux change. A current only exists in a closed conducting loop, so the law can only be applied directly where such a loop exists. If the loop is open, there is no current to speak of — but there is still an induced emf between the open ends. To find its direction you imagine the loop closed, work out which way the current would flow by Lenz's law, and read the emf's polarity off that imaginary current. So the law reaches the emf only indirectly, through the current it would drive. لینز کا قانون کرنٹ کی سمت کے بارے میں ہے، اور کرنٹ صرف بند لوپ میں بہتا ہے۔ کھلے لوپ میں ای ایم ایف تو ہوتی ہے مگر کرنٹ نہیں، اس لیے سمت معلوم کرنے کے لیے لوپ کو بند تصور کر کے کرنٹ کی سمت نکالی جاتی ہے اور اس سے ای ایم ایف کی سمت معلوم ہوتی ہے۔
No emf is induced — provided the loop stays completely inside the uniform field while it moves. The normal is parallel to the field, so the flux is at its maximum, BA. But moving the loop bodily through a uniform field changes neither B, nor the area, nor the angle, so the flux stays exactly constant, ΔΦ / Δt = 0, and Faraday's law gives no emf. ⚠️ The nuance matters: while the loop is entering or leaving the field, its flux is changing and an emf is induced. The book's question is about the steady motion through the field, where the answer is no. اگر لوپ مکمل طور پر یکساں میدان کے اندر حرکت کر رہا ہو تو فلکس تبدیل نہیں ہوتا، اس لیے کوئی ای ایم ایف پیدا نہیں ہوتی۔ البتہ میدان میں داخل ہوتے یا نکلتے وقت فلکس بدلتا ہے اور ای ایم ایف پیدا ہوتی ہے۔
No. The induced emf acts to oppose the change in flux, which is not the same thing. If the flux is increasing, the induced current's field opposes the increase — and in that case it does act to decrease the flux. But if the flux is decreasing, the induced current's field tries to maintain it, i.e. it acts to increase the flux back towards its old value. So the emf sometimes decreases the flux and sometimes increases it; what it always does is fight the change. نہیں۔ انڈیوسڈ ای ایم ایف فلکس کی تبدیلی کی مخالفت کرتی ہے: فلکس بڑھ رہا ہو تو اسے کم کرنے کی کوشش کرتی ہے، اور فلکس گھٹ رہا ہو تو اسے بڑھانے کی کوشش کرتی ہے۔
Current is produced because pushing the magnet in changes the magnetic flux through the solenoid, and a changing flux induces an emf (Faraday's law) which drives the current the galvanometer shows. The approaching pole in the figure is the north pole of the bar magnet. By Lenz's law the solenoid's left end must oppose the approach, so it becomes a north pole — like poles repel, and the repulsion fights the push. (The pole follows from Lenz's law applied to the book's figure; the book's text states the law, and this is its direct application.) مقناطیس اندر دھکیلنے سے سولینائیڈ سے گزرنے والا فلکس بدلتا ہے، اس لیے ای ایم ایف اور کرنٹ پیدا ہوتا ہے۔ قریب آنے والا قطب شمالی ہے، اس لیے لینز کے قانون سے سولینائیڈ کا بایاں سرا بھی شمالی قطب بنے گا تاکہ دھکے کی مخالفت ہو۔
No. As the magnet approaches the ring, the flux through the ring increases and an induced current flows so as to repel the approaching magnet; as it leaves, the induced current reverses and attracts it back. In both halves of the fall, Lenz's law gives a magnetic force opposing the motion — an upward drag on top of gravity. The net downward acceleration is therefore less than g. Once the magnet is far from the ring, the effect vanishes and it falls freely again. نہیں۔ مقناطیس کے قریب آنے پر رنگ میں ایسا کرنٹ پیدا ہوتا ہے جو اسے دھکیلتا ہے، اور دور جاتے وقت ایسا کرنٹ جو اسے کھینچتا ہے — دونوں صورتوں میں حرکت کی مخالفت ہوتی ہے، اس لیے اسراع g سے کم ہو جاتا ہے۔
(b) The slow-moving particle is deflected most. Two things point the same way: its radius r = mv / qB is smaller, so its path curves more tightly, and it spends longer inside the field region, giving the force more time to bend it. The fast particle does feel a larger force (F = qvB), but it also resists bending more (larger momentum) and crosses the field quickly — and the tighter circle of the slow particle wins. If "deflection" meant the size of the force alone, the fast particle would win, but for the bending of the path — which is what the question asks — it is the slow one. آہستہ ذرہ زیادہ مڑتا ہے، کیونکہ اس کا دائرہ r = mv / qB چھوٹا ہوتا ہے اور وہ میدان میں زیادہ دیر رہتا ہے۔
The electron suffers the greater deflection. Both enter at right angles with the same velocity and the same magnitude of charge, so the force qvB is the same size for both — but the radius of the circle is r = mv / qB, and the proton is about 1836 times heavier than the electron. The electron's circle is therefore about 1836 times tighter, so it bends far more sharply. Same force, very different mass, very different bending. الیکٹران زیادہ مڑتا ہے، کیونکہ r = mv / qB میں پروٹان کی کمیت تقریباً 1836 گنا زیادہ ہے، اس لیے الیکٹران کا دائرہ بہت چھوٹا بنتا ہے۔
Yes. A magnetic field is produced by moving charge, and a current is just a stream of moving charges. A single proton in motion is the smallest possible case of that: it carries its own magnetic field around with it, in circles about its line of motion, just as the chapter's opening explains that a current-carrying conductor is surrounded by a magnetic field. One proton's field is far too weak to measure with school apparatus, but it is real. جی ہاں۔ مقناطیسی میدان حرکت کرتے چارج سے پیدا ہوتا ہے، اور ایک حرکت کرتا پروٹان بھی اپنے گرد مقناطیسی میدان رکھتا ہے، اگرچہ وہ بہت کمزور ہوتا ہے۔
Flux is Φ = BA cos θ, so a field alone is not enough — the geometry has to cooperate. If the field lies in the plane of the coil, the angle between the field and the coil's normal is 90°, cos θ = 0, and the flux is zero even in a strong field. The flux is also zero if the coil's area is zero, and it is at its maximum only when the field runs along the normal. So a field is necessary for flux, but the flux also depends on the area and, crucially, on the orientation — which is why a coil can sit in a strong field and still thread no flux at all. فلکس Φ = BA cos θ ہے، اس لیے صرف میدان کافی نہیں: اگر میدان کوائل کی سطح میں ہو تو θ = 90° اور فلکس صفر ہو جاتا ہے، چاہے میدان کتنا ہی مضبوط ہو۔
No. The magnetic force on a charge is F = qvB sin θ, and a stationary charge has v = 0, so the force is zero no matter how strong the field between the poles is. A magnetic field only pushes charges that are moving across it; a charge at rest feels only electric forces, and there is no electric field here. Put the same charge in motion across the field and the force appears at once. نہیں، کیونکہ ساکن چارج کی رفتار v = 0 ہوتی ہے اور F = qvB sin θ صفر ہو جاتا ہے۔ مقناطیسی میدان صرف حرکت کرتے چارج پر قوت لگاتا ہے۔
Closing the switch builds up the coil's magnetic field from zero, so the flux through the ring above it increases rapidly. By Lenz's law the ring's induced current flows so as to oppose that increase — its field repels the coil's field — and the repulsion kicks the ring upward. ⚠️ The reversed-polarity part is not spelled out in a book passage, so here is the reasoning, flagged as derived: reversing the battery builds the field up in the opposite direction, so the flux change is reversed, the induced current reverses too, and the ring's field still opposes the change — so the ring still jumps upward. The kick's direction does not flip with the battery, because Lenz's law always opposes the change, whichever way the change runs. سوئچ بند ہونے سے کوائل کا میدان اچانک بنتا ہے، رنگ سے گزرنے والا فلکس بڑھتا ہے، اور لینز کے قانون سے رنگ میں ایسا کرنٹ پیدا ہوتا ہے جو اس اضافے کی مخالفت کرتا ہے — اس دھکیل سے رنگ اوپر اچھلتا ہے۔ بیٹری کی پولیرٹی الٹنے پر میدان مخالف سمت میں بنے گا، انڈیوسڈ کرنٹ بھی الٹ جائے گا، مگر مخالفت پھر بھی وہی رہے گی، اس لیے رنگ پھر بھی اوپر اچھلے گا۔
Rotate it about the x-axis. Start from the flux: the coil lies in the x–y plane, so its normal points along z, while B points along y — the two are perpendicular, so the initial flux is zero. Rotating about the y-axis (the field's own direction) leaves the normal in the x–z plane, still perpendicular to B, so the flux stays zero. Rotating about the z-axis (the normal itself) spins the coil in its own plane and the normal never moves, so the flux stays zero too. Only rotation about the x-axis tips the normal towards the y-direction, so the component of B along the normal — and hence the flux — changes, and an emf is induced. One axis works, two do nothing, and the reason is purely the flux geometry. صرف x-axis کے گرد گھمانے سے ای ایم ایف پیدا ہوگی۔ کوائل x–y سطح میں ہے، اس کا نارمل z سمت میں ہے اور B y سمت میں، اس لیے شروع میں فلکس صفر ہے۔ x-axis کے گرد گھمانے سے نارمل y کی طرف جھکتا ہے اور فلکس بدلتا ہے؛ y یا z-axis کے گرد گھمانے سے فلکس صفر ہی رہتا ہے۔
Yes. The induced emf depends on the rate of change of flux, Φ = BA cos θ, not on B or A separately. If the area grows while the field weakens in exactly the right proportion — say the area doubles while the field halves — the product BA stays constant, the flux does not change, and no emf appears. The same trick works with the angle: a growing area can be cancelled by a growing angle to the normal. Change both quantities, keep their product constant, and Faraday's law gives zero. جی ہاں۔ ای ایم ایف فلکس کی تبدیلی پر منحصر ہے، نہ کہ B یا A پر الگ الگ۔ اگر رقبہ دوگنا ہو جائے اور میدان آدھا رہ جائے تو BA مستقل رہے گا، فلکس نہیں بدلے گا اور کوئی ای ایم ایف پیدا نہیں ہوگی۔
No. The magnetic force F = q(v × B) is always perpendicular to the particle's velocity, and a force perpendicular to the motion does no work (work needs a component along the displacement). With no work done, the kinetic energy cannot change — the speed stays exactly the same, and only the direction of motion changes. That is why the particle circles at constant speed in a uniform field instead of speeding up or slowing down. (An electric field, by contrast, can do work and change kinetic energy, which is exactly how the velocity selector's plates are able to balance the magnetic force.) نہیں۔ مقناطیسی قوت ہمیشہ رفتار کے عموداً لگتی ہے، اس لیے وہ کوئی کام نہیں کرتی اور حرکی توانائی تبدیل نہیں ہوتی — صرف حرکت کی سمت بدلتی ہے، رفتار وہی رہتی ہے۔
Two forces act, and they do very different things. The magnetic force is zero: the proton moves exactly opposite to B, so θ = 180°, sin θ = 0, and qvB sin θ = 0 — motion along (or against) the field lines feels nothing. The electric force is qE, pointing along E, which is opposite to the proton's motion, so it acts as a brake: the proton slows down steadily, comes momentarily to rest, and then accelerates back along the direction of E. In short: it decelerates, stops, and returns the way the field points, with its speed growing again under the electric force alone. مقناطیسی قوت صفر ہوگی، کیونکہ پروٹان میدان کے عین مخالف سمت میں حرکت کر رہا ہے (θ = 180°)۔ برقی قوت qE حرکت کی مخالف سمت میں لگے گی، اس لیے پروٹان آہستہ ہوگا، لمحہ بھر رکے گا، پھر برقی میدان کی سمت میں واپس تیز ہوتا چلا جائے گا۔
Yes — and that reverse effect is the whole second half of this chapter. A current-carrying conductor moves because the field pushes its charges; turn it around, and moving the conductor drags its charges across the field, which pushes them along the conductor and piles them at one end. That separation is a motional emf, ε = LvB, and if the conductor is part of a closed circuit, a current flows. Motor action and generator action are the same physics read forwards and backwards: current plus field gives motion, motion plus field gives current. جی ہاں۔ موصل کو میدان میں حرکت دینے سے اس کے چارجز پر qvB قوت لگتی ہے، چارج ایک سرے پر جمع ہوتے ہیں اور موشنل ای ایم ایف ε = LvB پیدا ہوتی ہے؛ بند سرکٹ ہو تو کرنٹ بہنے لگتا ہے۔ موٹر اور جنریٹر یہی طبیعیات الٹی سمت میں ہیں۔
It comes from the mechanical work of whoever — or whatever — keeps pushing the rod to the right. Follow the energy: the induced current flows through the rod, and the field then exerts a magnetic force on that current-carrying rod pointing to the left, against the motion (Lenz's law again). If nothing pushed the rod, it would slow down and stop, and the bulb would go out. To keep it moving at constant speed, an external agent must push right with a force equal to that magnetic drag, doing work all the while. That mechanical work is converted into the electrical energy the bulb consumes (and then into light and heat). Energy is conserved throughout: no work in, no light out. یہ توانائی راڈ کو دھکیلنے والے کے مکینیکل کام سے آتی ہے۔ انڈیوسڈ کرنٹ کی وجہ سے راڈ پر میدان کی قوت حرکت کی مخالف سمت میں لگتی ہے، اس لیے راڈ کو مسلسل رفتار سے چلائے رکھنے کے لیے باہر سے کام کرنا پڑتا ہے — اور وہی کام برقی توانائی بن کر بلب کو روشن کرتا ہے۔
Shield it magnetically: enclose the instrument in a box made of a high-permeability soft magnetic material such as soft iron (in modern practice, mu-metal). The stray field lines prefer the easy path through the shielding material rather than through the air inside, so they are diverted around the box and the field inside drops to almost nothing. The instrument then sits in a nearly field-free pocket, protected from the stray fields outside. ⚠️ This answer is derived from the general behaviour of magnetic field lines in permeable materials; the book's chapter does not contain a dedicated passage on magnetic shielding, so it is flagged as derived rather than quoted. آلے کو نرم لوہے (یا میو میٹل) جیسے زیادہ نفوذ پذیری والے مقناطیسی مواد کے ڈبے میں بند کریں۔ آوارہ میدان کی لکیریں اس مواد سے گزرنا پسند کرتی ہیں، اس لیے وہ ڈبے کے گرد سے گزر جاتی ہیں اور اندر میدان تقریباً صفر رہ جاتا ہے۔
Magnetic flux, ΦB, is the total number of magnetic field lines passing through a given surface — a property of a particular surface in a field, measured in webers. Flux density, B, is the flux per unit area perpendicular to the field — a property of the field itself at a point, measured in teslas (Wb m−2). They are related by ΦB = BA cos θ, or, for a surface square-on to the field, ΦB = BA and B = ΦB / A. So flux tells you how much field threads one surface, flux density tells you how concentrated the field is, and the same field gives different fluxes through different surfaces — large through a big square-on loop, zero through a loop lying along the field lines. Flux is a scalar; flux density is a vector pointing along the field.
۱۔ مقناطیسی فلکس Φ کسی خاص سطح سے گزرنے والی میدان کی لکیروں کی کل تعداد ہے، اس کی اکائی ویبر ہے۔ ۲۔ فلکس کثافت B میدان کے عموداً فی اکائی رقبہ فلکس ہے، یعنی میدان کی اپنی خاصیت، اس کی اکائی ٹیسلا (Wb m−2) ہے۔ ۳۔ دونوں کا تعلق Φ = BA cos θ ہے، اور سطح میدان کے عموداً ہو تو B = Φ / A۔ ۴۔ ایک ہی میدان مختلف سطحوں سے مختلف فلکس دیتا ہے — بڑی عمودی سطح سے زیادہ، میدان کے ساتھ پڑی سطح سے صفر۔ ۵۔ فلکس اسکیلر ہے اور فلکس کثافت ویکٹر ہے۔
Experiment shows the force is directly proportional to three things: the current I in the conductor, the length L of the conductor inside the field, and the strength B of the external field. So F ∝ BIL, and in SI units the constant of proportionality is 1, giving F = BIL for a conductor perpendicular to the field (10.1). Rearranging defines the field strength: B = F / IL — the force on a 1 m wire carrying 1 A at right angles to the field. If the conductor makes an angle θ with the field, only the component of its length perpendicular to the field, L sin θ, contributes, so the general expression is F = BIL sin θ (10.2). In vector form, with L as a vector along the current, F = IL × B (10.3). The force is maximum (BIL) at θ = 90° and zero at θ = 0°, and its direction is given by Fleming's left-hand rule: first finger along B, second finger along I, thumb along F.
۱۔ تجربے سے معلوم ہوتا ہے کہ قوت کرنٹ I، لمبائی L اور میدان کی شدت B کے راست متناسب ہے: F ∝ BIL۔ ۲۔ SI اکائیوں میں تناسب کا مستقل 1 ہے، اس لیے میدان کے عموداً موصل کے لیے F = BIL۔ ۳۔ اس سے B کی تعریف ملتی ہے: B = F / IL۔ ۴۔ اگر موصل میدان سے θ زاویہ بنائے تو صرف عمودی جزو L sin θ کام کرتا ہے، اس لیے عام صورت F = BIL sin θ ہے۔ ۵۔ ویکٹر صورت: F = IL × B۔ ۶۔ قوت θ = 90° پر زیادہ سے زیادہ اور θ = 0° پر صفر ہوتی ہے، اور سمت فلیمنگ کے بائیں ہاتھ کے اصول سے معلوم ہوتی ہے۔
Faraday's law: the average emf induced in a coil of N loops is equal to the negative of the rate at which the magnetic flux through the coil is changing: ε = −N ΔΦB / Δt. It is proved in the book from motional emf: as a rod moves, the loop area grows, the flux grows, and substituting v = Δx / Δt into ε = vBL gives ε = −ΔΦ / Δt for one loop, times N for a coil. The key lesson is that the rate of change of flux matters, not the flux itself. Lenz's law: the direction of the induced current is always such that it opposes the change that causes the current. Push a magnet's north pole at a coil and the coil's near face becomes north to repel it; pull it away and the face becomes south to attract it. The minus sign in Faraday's law is Lenz's law in mathematical form, and it is required by conservation of energy — otherwise induced currents would feed their own motion forever. Factors affecting the emf (section 10.8): (1) rate of change of magnetic flux — faster change, bigger emf; (2) number of turns N — emf is proportional to N; (3) relative speed of coil or conductor through the field — faster motion, faster flux change, bigger emf.
۱۔ فیراڈے کا قانون: N چکروں والی کوائل میں اوسط انڈیوسڈ ای ایم ایف فلکس کی تبدیلی کی شرح کے منفی کے برابر ہے: ε = −N ΔΦ / Δt۔ اہم بات یہ ہے کہ فلکس کی مقدار نہیں، اس کی تبدیلی کی شرح ای ایم ایف بناتی ہے۔ ۲۔ لینز کا قانون: انڈیوسڈ کرنٹ کی سمت ہمیشہ اس تبدیلی کی مخالفت کرتی ہے جس سے کرنٹ پیدا ہوا ہو — مقناطیس دھکیلیں تو کوائل کا سرا شمالی بن کر دھکے کی مخالفت کرتا ہے، کھینچیں تو جنوبی بن کر کھنچاؤ کی مخالفت کرتا ہے۔ یہی قانون توانائی کے تحفظ کا دوسرا روپ ہے۔ ۳۔ ای ایم ایف کو متاثر کرنے والے عوامل: فلکس کی تبدیلی کی شرح، چکروں کی تعداد N، اور میدان میں نسبتی رفتار — تینوں بڑھیں تو ای ایم ایف بڑھتی ہے۔
Start from the wire: a conductor of length L carrying N charges q in time t carries current I = Nq / t, and since the charges drift at speed v across length L in time t = L / v, this becomes I = Nqv / L. A wire in a uniform field B feels F = IL × B, so the whole conductor feels F = (Nqv / L)(L × B) = Nq(v × B) — N times the force on one particle. Dividing by N, the force on a single particle is F = q(v × B) (10.13), with magnitude F = qvB sin θ (10.14), where θ is the angle between v and B. It is maximum, qvB, when the particle crosses the field at right angles, and zero when it moves along the field. Its direction follows Fleming's left-hand rule (or the right-hand rule for the cross product), reversed for negative charges. Because the force is always perpendicular to v, it changes only the direction of motion: at right angles to a uniform field the particle circles with radius r = mv / qB, the magnetic force supplying the centripetal force.
۱۔ تار میں کرنٹ I = Nq / t ہے، اور t = L / v رکھنے سے I = Nqv / L بنتا ہے۔ ۲۔ تار پر قوت F = IL × B ہے، اس لیے پوری تار پر F = Nq(v × B)۔ ۳۔ ایک ذرے پر قوت F = q(v × B)، اور مقدار F = qvB sin θ۔ ۴۔ قوت θ = 90° پر زیادہ سے زیادہ (qvB) اور θ = 0° پر صفر ہوتی ہے۔ ۵۔ سمت فلیمنگ کے اصول سے، اور منفی چارج کے لیے الٹی ہوتی ہے۔ ۶۔ قوت ہمیشہ رفتار کے عموداً ہوتی ہے، اس لیے ذرہ یکساں میدان میں عموداً داخل ہو تو r = mv / qB رداس کے دائرے میں گھومتا ہے۔
A velocity selector is a device that lets through charged particles of one chosen speed and deflects all others. It consists of a cylindrical tube inside a magnetic field B, containing a parallel-plate capacitor that makes a uniform electric field E perpendicular to B. A positive particle entering the tube feels a magnetic force qvB in one direction and an electric force qE in the opposite direction. For most speeds one force wins and the particle bends away from the exit. But when qvB = qE, the forces cancel exactly, the net force is zero, and by Newton's first law the particle keeps its velocity and travels straight through to the far end. Cancelling q gives the selected speed v = E / B — independent of the particle's charge and mass, so the device selects purely by speed. Particles faster or slower than v are deflected and do not reach the exit. The book's examples show both uses: with E = 300 N C−1 and B = 0.20 T only the 1500 m s−1 alpha particles pass; and a selector followed by a magnetic circle lets you measure q / m, since v = E / B first and then q / m = v / Br = E / (B2r).
۱۔ ویلاسٹی سلیکٹر ایک ایسا آلہ ہے جو صرف ایک خاص رفتار والے ذرات کو گزرنے دیتا ہے اور باقی سب کو موڑ دیتا ہے۔ ۲۔ اس میں مقناطیسی میدان B کے اندر ایک متوازی پلیٹوں والا کیپیسٹر ہوتا ہے جو برقی میدان E بناتا ہے، جو B کے عموداً ہوتا ہے۔ ۳۔ ذرے پر مقناطیسی قوت qvB ایک طرف اور برقی قوت qE دوسری طرف لگتی ہے۔ ۴۔ جب qvB = qE ہو تو دونوں قوتیں برابر ہو جاتی ہیں، خالص قوت صفر ہوتی ہے اور ذرہ سیدھا گزر جاتا ہے۔ ۵۔ اس سے منتخب رفتار v = E / B ملتی ہے، جو چارج اور کمیت پر منحصر نہیں۔ ۶۔ تیز یا آہستہ ذرات مڑ جاتے ہیں اور باہر نہیں نکلتے۔
A ferrofluid is a colloidal suspension of magnetic nanoparticles — usually iron oxide, about 10 nm across — in a carrier fluid such as oil or water, each particle coated with a surfactant that stops the particles clumping and keeps them evenly dispersed; their small size and constant motion stop them settling. With no magnet nearby it behaves like an ordinary liquid. Bring a magnet near and the particles are temporarily magnetised: they line up in chains along the field lines, and the tug-of-war between magnetic force, surface tension and gravity raises the surface into stiff spikes wherever the magnetic force wins. Remove the magnet and the particles demagnetise, the spikes collapse, and the fluid is a liquid again — it never stays magnetic by itself. The book's home experiment makes the same fluid by stirring laser-printer toner (about 40% nanometre iron oxide) into cooking oil: shaken, it flows like a liquid; with a magnet at the bottle's side, it jumps to the magnet and grows spikes. Its uses follow from that switchable magnetism: sealing rotating hard-drive shafts, cooling and damping loudspeaker voice coils, steering drugs to tumours with an external magnet, MRI contrast, and controlling liquid flow with a shaped field.
۱۔ فیرو فلوئڈ تیل یا پانی جیسے مائع میں آئرن آکسائیڈ کے تقریباً 10 نینومیٹر جسامت کے ذرات کی معلقیت ہے، ہر ذرے پر سرفیکٹنٹ کی تہہ ہوتی ہے جو انہیں جمنے اور نیچے بیٹھنے سے روکتی ہے۔ ۲۔ مقناطیس کے بغیر یہ عام مائع کی طرح برتاؤ کرتا ہے۔ ۳۔ مقناطیس قریب لائیں تو ذرات عارضی طور پر مقناطیسی ہو کر میدان کی لکیروں کے ساتھ زنجیروں میں لگ جاتے ہیں، اور مقناطیسی قوت، سطحی تناؤ اور کشش ثقل کی کشمکش سے سطح پر سپائکس بن جاتے ہیں۔ ۴۔ مقناطیس ہٹائیں تو ذرات غیر مقناطیسی ہو جاتے ہیں اور مائع پھر عام مائع بن جاتا ہے۔ ۵۔ استعمالات: ہارڈ ڈرائیو کی سیل، لاؤڈ اسپیکر کی ٹھنڈک اور آواز کی بہتری، مقناطیس سے دوا کو ٹیومر تک پہنچانا، ایم آر آئی کنٹراسٹ، اور مائع کے بہاؤ کا کنٹرول۔
Step 1 — first projection: perpendicular, so sin 90° = 1 and F = q × 1500 × B. Step 2 — second projection: F = q × 2000 × B × sin θ. Step 3 — the forces are equal, and q and B are the same both times, so cancel them: 1500 = 2000 sin θ, giving sin θ = 1500 / 2000 = 0.75. Step 4 — θ = sin−1(0.75) = 48.6°, which rounds to θ = 49°, matching the book's answer. (The supplementary angle 180° − 49° = 131° would give the same sine, but the acute angle is the intended answer.) پہلی بار F = q × 1500 × B، دوسری بار F = q × 2000 × B × sin θ۔ برابر رکھنے سے sin θ = 1500 / 2000 = 0.75، اس لیے θ = 48.6° ≈ 49°۔
Step 1 — the accelerating voltage gives the electron kinetic energy: (1/2)mv2 = eV, so v = √(2eV / m) = √(2 × 1.6 × 10−19 × 15000 / 9.11 × 10−31) = √(5.27 × 1015) = 7.26 × 107 m s−1. Step 2 — the force is maximum when the electron crosses the field at right angles: Fmax = evB = 1.6 × 10−19 × 7.26 × 107 × 0.35 = 4.1 × 10−12 N, matching the book's answer. پہلے v = √(2eV / m) = 7.26 × 10⁷ m s−1، پھر زیادہ سے زیادہ قوت F = evB = 4.1 × 10−12 N۔
Step 1 — area of one turn: A = 0.15 × 0.15 = 2.25 × 10−2 m2. Step 2 — initial flux linkage: the plane is perpendicular to the field, so the normal is along the field, θ = 0°, and Φi = NBA = 60 × 0.8 × 2.25 × 10−2 = 1.08 Wb. Step 3 — after turning 30°, the angle between the normal and the field is 30°: Φf = NBA cos 30° = 1.08 × 0.866 = 0.935 Wb. Step 4 — change in linkage: ΔΦ = 1.08 − 0.935 = 0.145 Wb. Step 5 — average emf: ε = ΔΦ / Δt = 0.145 / 2 = 0.0725 ≈ 0.07 V, matching the book's answer. A = 0.15 × 0.15 = 2.25 × 10−2 m²، شروع میں Φ = 60 × 0.8 × 2.25 × 10−2 = 1.08 Wb، 30° گھمانے کے بعد Φ = 1.08 × 0.866 = 0.935 Wb، اس لیے ε = 0.145 / 2 ≈ 0.07 V۔
Motional emf is directly proportional to the field strength when length and speed are unchanged: ε ∝ B. So ε2 / ε1 = B2 / B1, giving B2 = B1 × (ε2 / ε1) = 0.2 × (2.4 / 0.8) = 0.2 × 3 = 0.6 T. Tripling the emf needs tripling the field — a one-line ratio question, as long as you spot that L and v are held constant. ε ∝ B ہے، اس لیے B₂ = 0.2 × (2.4 / 0.8) = 0.6 T۔
Step 1 — area of the ring: A = πr2 = 3.14 × (0.04)2 = 5.03 × 10−3 m2. Step 2 — change in field: ΔB = 0.4 − 0.2 = 0.2 T. Step 3 — induced emf: ε = A × ΔB / Δt = 5.03 × 10−3 × 0.2 / (5 × 10−3) = 5.03 × 10−3 × 40 = 0.201 V. Step 4 — current: I = ε / R = 0.201 / (1.0 × 10−3) = 201 A, matching the book's answer. The huge current is no surprise: the resistance is only a milliohm, so even a fifth of a volt drives hundreds of amperes. A = π × (0.04)² = 5.03 × 10−3 m²، ε = 5.03 × 10−3 × 0.2 / (5 × 10−3) = 0.201 V، اس لیے I = 0.201 / (1.0 × 10−3) = 201 A۔
Step 1 — convert the area: A = 35 cm2 = 35 × 10−4 m2 = 3.5 × 10−3 m2. Step 2 — initial flux linkage: Φi = NBA = 10 × 0.5 × 3.5 × 10−3 = 1.75 × 10−2 Wb. Step 3 — once the coil is out of the field the flux is zero, so ΔΦ = 1.75 × 10−2 Wb. Step 4 — emf: ε = ΔΦ / Δt = 1.75 × 10−2 / 1.0 = 1.75 × 10−2 V, matching the book's answer. A = 35 × 10−4 = 3.5 × 10−3 m²، Φ = 10 × 0.5 × 3.5 × 10−3 = 1.75 × 10−2 Wb، اس لیے ε = 1.75 × 10−2 / 1.0 = 1.75 × 10−2 V۔
Step 1 — speed from the accelerating voltage: (1/2)mv2 = eV, so v = √(2eV / m) = √(2 × 1.6 × 10−19 × 6 × 105 / 1.67 × 10−27) = √(1.15 × 1014) = 1.07 × 107 m s−1. Step 2 — radius in the field: r = mv / (eB) = (1.67 × 10−27 × 1.07 × 107) / (1.6 × 10−19 × 1.0) = 1.79 × 10−20 / 1.6 × 10−19 = 0.112 m = 11.2 cm, matching the book's answer. Note the field unit: 1 weber m−2 is 1 tesla, so no conversion is needed. پہلے v = √(2eV / m) = 1.07 × 10⁷ m s−1، پھر r = mv / eB = 11.2 cm۔ (1 Wb m−2 = 1 T ہے۔)
At an angle to the field the proton follows a helix, and the circular part of that helix is governed by the component of velocity perpendicular to the field: v⊥ = v sin 45° = 104 × 0.707 = 7.07 × 103 m s−1. (The parallel component carries the proton along the field lines and does not affect the circle's radius.) Then r = mv⊥ / (qB) = (1.67 × 10−27 × 7.07 × 103) / (1.6 × 10−19 × 0.3) = 1.18 × 10−23 / 4.8 × 10−20 = 2.46 × 10−4 m, matching the book's answer. The trap here is using the full velocity instead of its perpendicular component. دائرے کے لیے صرف عمودی جزو کام آتا ہے: v⊥ = 10⁴ × sin 45° = 7.07 × 10³ m s−1، اس لیے r = mv⊥ / qB = 2.46 × 10−4 m۔
Motional emf needs all three directions to be mutually perpendicular: the emf along a rod is (v × B) · L, and it is only non-zero when the rod's length has a component along v × B. Only one of the three rods satisfies that: it moves sideways across the field, cutting field lines, so its emf is ε = vBL = 2.5 × 0.5 × 0.60 = 0.75 V. The force q(v × B) on its positive charges points towards end a, so end a is positive. The other two rods give zero: one moves parallel to the field (v × B = 0), and the other moves along its own length, so (v × B) is perpendicular to the rod and no charge separation builds up along it. ⚠️ The book's figure is small and its rod labels are hard to read in the scan, so the identification of which rod is L1 follows the book's printed answer key: L1 is the sideways-moving rod, and the other two are the zero cases. The physics — one rod cuts the field, two do not — is unambiguous even where the labelling is small. صرف وہ راڈ ای ایم ایف بناتی ہے جو میدان کو کاٹتے ہوئے ایک طرف حرکت کرتی ہے: ε = 2.5 × 0.5 × 0.60 = 0.75 V، اور q(v × B) کی سمت سے سرا a مثبت ہوتا ہے۔ باقی دو راڈوں میں ایک میدان کے متوازی حرکت کرتی ہے اور دوسری اپنی ہی لمبائی کی سمت میں، اس لیے دونوں میں ای ایم ایف صفر ہے۔
With length and speed unchanged, ε ∝ B, so B2 = B1 × (ε2 / ε1) = 0.4 × (1.5 / 0.5) = 0.4 × 3 = 1.2 T. The same ratio argument as Numerical 10.4, with different numbers — spot the proportionality and the question is one line. ε ∝ B ہے، اس لیے B₂ = 0.4 × (1.5 / 0.5) = 1.2 T۔
Undeflected means the electric and magnetic forces balance: qE = qvB, so E = vB = 4.96 × 104 × 0.114 = 5654.4 ≈ 5.65 × 103 N C−1, matching the book's answer. This is the velocity-selector condition v = E / B turned around to solve for E. بغیر مڑے گزرنے کے لیے qE = qvB، اس لیے E = vB = 4.96 × 10⁴ × 0.114 = 5.65 × 10³ N C−1۔
Step 1 — magnitude: F = BIL with the conductor perpendicular to the field, so I = F / (BL) = 1 / (0.5 × 2) = 1 A. Step 2 — direction: the force points into the page and the field points to the right in the figure. By Fleming's left-hand rule (first finger along B to the right, thumb into the page), the second finger — the current — points upwards along the conductor, i.e. from Q at the bottom to P at the top. So the current is 1 A, from Q to P, matching the book's answer. I = F / BL = 1 / (0.5 × 2) = 1 A۔ فلیمنگ کے اصول سے کرنٹ اوپر کی طرف، یعنی Q سے P کی طرف بہتا ہے۔
This chapter is really two stories joined at the middle. The first story is fields pushing charges: a current-carrying wire feels F = BIL sin θ, a single charge feels F = qvB sin θ and circles with radius r = mv / qB, and a velocity selector uses the balance qE = qvB to pick out one speed, v = E / B. The second story is changing fields making voltages: Faraday's law ε = −N ΔΦ / Δt, with Lenz's law fixing the direction by opposing the change, and three factors — rate of flux change, number of turns, relative speed — setting the size. The chapter then cashes those ideas out in two real devices: ferrofluids, where a liquid borrows magnetism for as long as a field is present, and the seismometer, where ground motion becomes an electrical signal through induction.
Revision guidance: learn the five formulas cold — F = BIL sin θ, F = qvB sin θ, r = mv / qB, v = E / B, and ε = −N ΔΦ / Δt — and the two zero cases that examiners love: force zero when motion is along the field, and emf zero when flux is constant. Then practise the numericals in order: 10.1 (the sine ratio), 10.2 and 10.7 (voltage to speed to force or radius), 10.3, 10.5 and 10.6 (flux linkage changes), 10.4 and 10.10 (the ε ∝ B ratio), 10.8 (the perpendicular component), 10.9 (which rod actually cuts the field), 10.11 (the selector turned around) and 10.12 (Fleming's rule backwards). For theory questions, the Lenz's-law energy argument (section 10.7) and the velocity selector working (section 10.5) are the two answers most likely to be asked in full.
اس باب میں دو کہانیاں ہیں۔ پہلی کہانی میدانوں کا چارجز کو دھکیلنا ہے: تار پر F = BIL sin θ، ایک ذرے پر F = qvB sin θ اور دائرے کا رداس r = mv / qB، اور ویلاسٹی سلیکٹر میں qE = qvB سے منتخب رفتار v = E / B۔ دوسری کہانی بدلتے میدانوں سے وولٹیج بننا ہے: فیراڈے کا قانون ε = −N ΔΦ / Δt، لینز کا قانون سمت طے کرتا ہے، اور تین عوامل — فلکس کی تبدیلی کی شرح، چکروں کی تعداد، نسبتی رفتار — مقدار طے کرتے ہیں۔ آخر میں دو حقیقی آلات ہیں: فیرو فلوئڈ، جو میدان موجود ہو تو مقناطیس بن جاتا ہے، اور سیسمومیٹر، جو زمین کی حرکت کو انڈکشن سے برقی سگنل بناتا ہے۔
نظرثانی کی رہنمائی: پانچ فارمولے زبانی یاد کریں — F = BIL sin θ، F = qvB sin θ، r = mv / qB، v = E / B، اور ε = −N ΔΦ / Δt — اور دو صفر والی صورتیں: میدان کے ساتھ حرکت پر قوت صفر، اور ساکن فلکس پر ای ایم ایف صفر۔ عددی سوالات ترتیب سے مشق کریں: 10.1 (سائن کا تناسب)، 10.2 اور 10.7 (وولٹیج سے رفتار)، 10.3، 10.5 اور 10.6 (فلکس لنکیج کی تبدیلی)، 10.4 اور 10.10 (ε ∝ B کا تناسب)، 10.8 (عمودی جزو)، 10.9 (کون سی راڈ میدان کاٹتی ہے)، 10.11 (سلیکٹر الٹا) اور 10.12 (فلیمنگ کا اصول الٹا)۔ نظریاتی سوالات میں لینز کے قانون کی توانائی والی دلیل اور ویلاسٹی سلیکٹر کا طریقہ کار سب سے زیادہ پوچھے جانے والے جواب ہیں۔
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