NCERT Exemplar Electromagnetic Induction - Physics Class 12 - NEET PDF

MULTIPLE CHOICE QUESTIONS

Q.1. A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by 

where B_o is constant. The magnitude of flux passing through the square is

(a) 2 B_o L²  Wb.

(b) 3 B_o L²  Wb.

(c) 4 B_o L²  Wb.

(d) B_o L² Wb.

Ans: (c)

Explanation: Magnetic flux through a surface is Φ = ∫ B·dA, where dA is the area element and the integrand is the component of B normal to the surface. For the square lying in the x-y plane the area vector is along +z. Evaluating the surface integral of the z-component of the given field over the square of side L gives the total flux as 4B₀L² Wb. The intermediate expressions and stepwise integral are shown in the figures below.

Q.2. A loop, made of straight edges has six corners at A(0,0,0), B(L,O,0) C(L ,L ,0) , D (0,L,0) E (0,L,L ) a n d F (0,0,L). A magnetic field 

 is present in the region. The flux passing through the loop ABCDEFA (in that order) is
(a) B_o L² Wb.
(b) 2 B_o L² Wb.
(c) √2 B_o L² Wb.
(d) 4 B_o L²  Wb.

Ans: (b)

Explanation: The loop ABCDEFA can be regarded as two planar loops: ABCDA lying in the x-y plane and ADEFA lying in the y-z plane. Each planar part contributes flux equal to the appropriate component of B times its area. The total flux is the algebraic sum of fluxes through these two orthogonal faces. Evaluating the dot products of the area vectors with the given B field and adding the two contributions yields 2B₀L² Wb. The area vectors and the intermediate flux expressions are displayed in the figures below.

Q.3. A cylindrical bar magnet is rotated about its axis (Fig 6.1). A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then
(a) A direct current flows in the ammeter A.
(b) No current flows through the ammeter A.
(c) An alternating sinusoidal current flows through the ammeter A with a time period T=2π/ω.
(d) A time varying non- sinusoidal current flows through the ammeter A

Ans: (b)

Explanation: Electromotive force is induced only when the magnetic flux linked with a circuit changes. Rotating the cylindrical magnet about its own symmetry axis does not change the magnetic flux through the circuit formed by the axis-to-surface contact because the field pattern around the axis remains the same. Hence no change of flux occurs and no induced emf is produced; therefore no current flows and the ammeter shows no deflection.

Q.4. There are two coils A and B as shown in Fig 6.2. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counterclockwise. B is kept stationary when A moves. We can infer that
(a) There is a constant current in the clockwise direction in A.
(b) There is a varying current in A.
(c) There is no current in A.
(d) There is a constant current in the counterclockwise direction in A.

Ans: (d)

Explanation: The changing flux through coil B is caused by the movement of coil A. The induced emf and current in B stop as soon as coil A stops moving, which implies the flux from A is no longer changing. That can happen only if coil A carries a steady (constant) current rather than a varying one. Since the given current direction in A is counterclockwise, the correct inference is that there is a constant counterclockwise current in A.

Q.5. Same as problem 4 except the coil A is made to rotate about a vertical axis (Fig 6.3). No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counterclockwise and the coil A is as shown at this instant, t = 0, is
(a) Constant current clockwise.
(b) Varying current clockwise.
(c) Varying current counterclockwise.
(d) Constant current counterclockwise.

Ans: (a)

Explanation: Lenz's law determines the direction of the induced current so as to oppose the change producing it. At t = 0, coil B has a counterclockwise current which produces a particular magnetic effect at coil A. When coil A rotates about the vertical axis, the relative flux linked with it changes in a manner that requires coil A to produce a magnetic effect opposing that change. Applying Lenz's rule to the geometry given shows that the required current in A is a steady (constant) clockwise current at that instant.

Q.6. The self inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as
(a) l and A increase.
(b) l decreases and A increases.
(c) l increases and  A decreases.
(d) both l and A decrease.

Ans: (b)

Explanation: For a long solenoid with N turns and length l the self inductance is L = μ_rμ₀N²A/l. Thus L is directly proportional to the cross-sectional area A and inversely proportional to length l. Therefore L increases when l decreases and A increases. The permeability μ_r of the core also increases L when the solenoid contains a magnetic core.

Important point: Self and mutual inductances depend on geometry and on the permeability of the medium.

Q.7. A metal plate is getting heated. It can be because

(a) A direct current is passing through the plate.

(b) It is placed in a time varying magnetic field.

(c) It is placed in a space varying magnetic field, but does not vary with time. 

(d) A current (either direct or alternating) is passing through the plate.

Ans: (a, b, c)

Explanation: Heating of a metal plate can occur by ordinary Joule heating when a direct or alternating current passes through it. In addition, a time-varying magnetic flux through the bulk conductor induces circulating eddy currents; these currents dissipate energy as heat even if no external source is connected. A spatially varying but time-independent magnetic field does not induce eddy currents by itself, but when there is relative motion (or a change in geometry) it can produce currents. The listed key facts about eddy currents - their circulation, naming after Foucault, and reduction by lamination - explain how a conducting plate may be heated under the given conditions.

Q.8. An e.m.f is produced in a coil, which is not connected to an external voltage source. This can be due to
(a) The coil being in a time varying magnetic field.
(b) The coil moving in a time varying magnetic field.
(c) The coil moving in a constant magnetic field.
(d) The coil is stationary in external spatially varying magnetic field, which does not change with time.

Ans: (a, b, c)

Explanation: An emf is induced in a closed conductor when the magnetic flux linked with it changes. This change can be caused by (a) a time-varying magnetic field at fixed coil position, (b) motion of the coil through a magnetic field that varies with time or position, or (c) motion of the coil in a spatially non-uniform but time-independent magnetic field so that the enclosed flux changes. A coil that is stationary in a purely spatially varying field with no time dependence does not experience a changing flux and so no emf is induced.

Q.9. The mutual inductance M₁₂ of coil 1 with respect to coil 2
(a) Increases when they are brought nearer.
(b) Depends on the current passing through the coils.
(c) Increases when one of them is rotated about an axis.
(d) Is the same as M₂₁ of coil 2 with respect to coil 1.

Ans: (a, d)

Explanation: Mutual inductance is a purely geometric property of two circuits (and the medium) and does not depend on the instantaneous current values; it depends on the relative positions and orientations of the coils. Bringing coils closer increases the fraction of flux of one coil that links the other, so M increases. Mutual inductance is symmetric: M₁₂ = M₂₁. Rotating a coil may change M if the relative orientation of the coils changes, but rotation about an axis that preserves the linking will not change it.

Q.10. A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because
(a) The magnetic field is constant.
(b) The magnetic field is in the same plane as the circular coil and it may or may not vary.

(c) The magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably.
(d) There is a constant magnetic field in the perpendicular (to the plane of the coil) direction.

Ans: (b, c)

Explanation: No emf is induced if the magnetic flux through the coil does not change. If the magnetic field lies in the plane of the coil, its normal component is zero and expansion of the coil does not change the flux (option b). Alternatively, the perpendicular component of B could decrease in time precisely so that the product B⊥·Area remains constant as the coil expands - then the flux stays constant and no emf appears (option c). A constant perpendicular field with increasing area would change the flux, so (d) cannot by itself explain zero emf.

VERY SHORT ANSWER TYPE QUESTIONS

Q.11. Consider a magnet surrounded by a wire with an on/off switch S (Fig 6.4). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain.

Ans: Key concept: Magnetic flux through a circuit is Φ = ∫ B·dA; an induced emf appears only when this flux changes.

If the bar magnet and the coil are both stationary and the switch merely closes the electrical circuit, there is no change in magnetic flux through the coil (neither B, nor area, nor the relative orientation changes). Therefore no induced emf is produced and no current flows as a result of simply closing the switch. Closing the switch only completes the circuit, but without a changing flux there is no induced current.

Q.12. A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain.

Ans: Key concept: Lenz's law and magnetic flux leakage.
The stretching separates turns and allows magnetic flux to leak between neighbouring turns, thereby reducing the effective flux linkage for each turn. The induced emf that would oppose changes in flux must, by Lenz's law, act to oppose this reduction. To oppose the flux decrease the circuit tends to increase current. Practically, as the coil is stretched the effective inductance decreases and so the steady current supplied by the DC source will increase (provided the source maintains the same voltage). Thus the current increases.

Q.13. A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain.

Ans: When the iron core is inserted, the magnetic field and flux inside the solenoid increase because the iron increases the permeability. By Lenz's law, any induced emf will oppose the change that produces it; the induced effect acts so as to reduce the flux increase. Thus the net effect is to reduce the circuit current slightly while the core is being inserted. If the source is an ideal voltage source, after insertion the steady current will be larger (since inductance rises); however the instantaneous induced emf during insertion opposes the change, producing a transient decrease. The usual classroom statement: insertion of an iron core increases flux and the induced response opposes that change during insertion, so the current momentarily decreases during insertion.

Q.14. Consider a metal ring kept on top of a fixed solenoid (say on a cardboard) (Fig 6.5). The centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain

Ans: When current in the solenoid is suddenly switched on, the magnetic flux through the ring increases rapidly. By Lenz's law an induced current appears in the ring whose magnetic field opposes the increase. The induced current produces a magnetic pole on the ring that repels the solenoid's near pole; the resulting repulsive magnetic force lifts the ring and it jumps upward. The direction of induced current is such that it generates this repulsion.

Q.15. Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current I (see Fig 6.5). The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring?

Ans: When the solenoid current is switched off the magnetic flux linking the ring decreases. By Lenz's law an induced current appears in the ring that tends to maintain the flux; this induced current produces a magnetic field which attracts the solenoid, so the ring experiences a downward magnetic force. If the ring is free to move it will be attracted downward toward the solenoid; if it sits on cardboard, the attraction is present but the ring cannot move downward due to the support.

Q.16. Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8cm is dropped through the pipe, it takes more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain

Ans: Key concept: Lenz's law and eddy currents.
As the magnet falls through the conducting pipe, the changing magnetic flux through portions of the pipe induces strong eddy currents. These eddy currents produce magnetic fields that oppose the motion of the magnet, producing a retarding magnetic force and reducing the magnet's acceleration. An unmagnetised iron bar does not produce the same changing flux pattern and so experiences negligible eddy-current braking; it falls faster. Therefore the magnet takes more time to pass through the pipe than the unmagnetised bar.

SHORT ANSWER TYPE QUESTIONS

Q.17. A magnetic field in a certain region is given by B = B₀ cos (ωt) 

and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (see Fig 6.6) . Find the magnitude and the direction of the current at (a, 0, 0) at t =π /2ω , t =π / ω and t = 3π /2ω .

Ans: Key concept: Faraday's law: induced emf ε = -dΦ/dt, and Ohm's law i = ε/R.
Flux through the coil is Φ = B(t)·A = B₀cos(ωt)·πa².
Therefore ε = -dΦ/dt = B₀ωπa² sin(ωt). The magnitude of induced current is |i| = (B₀ωπa²/R) |sin(ωt)| and its sense (direction) follows from Lenz's law: it is such as to oppose the change of flux.

Evaluate at the given times:

(i) t = π/(2ω): sin(ωt) = sin(π/2) = 1 → i = (B₀ωπa²/R) (direction given by Lenz's rule opposing the decreasing/increasing flux at that instant).

(ii) t = π/ω: sin(ωt) = sin(π) = 0 → i = 0.

(iii) t = 3π/(2ω): sin(ωt) = sin(3π/2) = -1 → i = -(B₀ωπa²/R) (direction opposite to case (i)).

The figures below indicate the flux, the sign of sin(ωt) at each instant and the resulting direction of induced current.

Q.18. Consider a closed loop C in a magnetic field (Fig 6.7). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula

 + ... .
 Now if we chose two different surfaces S₁ and S₂ having C as their edge, would we get the same answer for flux. Justify your answer.

Ans: Yes. Magnetic field lines are continuous and have no beginning or end (there are no magnetic monopoles). Therefore the net number of field lines passing through any two surfaces that share the same closed boundary C must be equal. Consequently the magnetic flux computed by Φ = ∫ B·dA is independent of the particular surface chosen, provided the boundary is the same. The illustrations below make this continuity argument clear.

Important point: Magnetic field lines can neither be created nor destroyed; this property underlies the surface-independence of flux for a given closed edge.

Q.19. Find the current in the wire for the configuration shown in Fig 6.8. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper . θ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

Ans: Key concept: Motional emf: emf induced across a conductor of length ℓ moving with velocity v in a magnetic field B is ε = Bℓv (projected component perpendicular to v and B included).

The induced electric field along the conductor is E = vB. For a rod PQ of length equal to the separation d moving at angle θ, the effective perpendicular component is vB sin θ. Thus the motional emf across PQ is ε = B v (length PQ) sin θ. If the closed circuit has total resistance R, the induced current is i = ε/R = (B v · PQ · sin θ)/R. The stepwise geometry and signs are shown in the figures below.

Q.20. A (current vs time) graph of the current passing through a solenoid is shown in Fig 6.9. For which time is the back electromotive force (u) a maximum. If the back emf at t = 3s is e, find the back emf at t = 7 s, 15s and 40s. OA, AB and BC are straight line segments.

Ans: The back emf ε = -L (dI/dt) is maximum where the magnitude of the slope dI/dt is largest. From the graph this occurs in the AB segment (between 5 s and 10 s). Given that at t = 3 s (in OA) the back emf is e, determine the slopes:

The slope in OA is (change in I)/(change in t) = (1 A - 0)/(5 s - 0) = 1/5 A s^-1. Since ε ∝ -dI/dt, at t = 3 s (OA) ε = -L(1/5) = e ⇒ L(1/5) = -e in sign convention used; using magnitudes, L(1/5) = |e|.

In AB the slope is three times that of OA (graph scale), so at t = 7 s the back emf magnitude is three times that at 3 s but with sign determined by the direction of current change. Thus ε(7 s) = -3e. In BC (10 s < t="" />< 30="" s)="" the="" slope="" is="" -1/10="" of="" oa="" (slower="" decrease),="" giving="" ε(15="" s)="e/2" (with="" sign="" depending="" on="" decreasing="" current).="" for="" t="" /> 30 s the current is constant so ε(40 s) = 0. The figures below show the segment slopes and the algebraic relations used.

Q.21. There are two coils A and B separated by some distance. If a current of 2 A flows through A, a magnetic flux of 10^-2 Wb passes through B (no current through B). If no current passes through A and a current of 1 A passes through B, what is the flux through A?

Ans: Let mutual inductance be M. Flux through B due to current in A is Φ_B = M·I_A = 10^-2 Wb when I_A = 2 A. Thus M = 10^-2/2 = 5×10^-3 H. By symmetry Φ_A = M·I_B so with I_B = 1 A the flux through A is Φ_A = 5×10^-3 Wb (i.e., 5 mWb).

LONG ANSWER TYPE QUESTIONS

Q.22. A magnetic field

sin (ωt)

covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (Fig. 6.10). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?

Ans: Key concept: The total emf has two contributions: the motional emf Blv due to the motion of AB in the magnetic field and the emf induced in the closed loop by the time variation of the magnetic field (Faraday's law).

Take AB at x(t) = vt at time t. Motional emf across AB: ε_motion = B(t)·l·v = B₀sin(ωt)·d·v. Emf due to changing B through the area swept by the loop is ε_flux = - d/dt ∫ B·dA over the loop area (the figure shows the integration domain). The net emf is the sum of these two contributions (signs set by chosen orientation). The current around the loop is i = (ε_motion + ε_flux)/R. The magnetic force on the rod is F = i d B (since force = iℓ×B). To keep the rod moving at constant velocity an external force equal and opposite to the magnetic drag is required: F_ext = -i d B directed along +x. The detailed algebra and the equivalent electrical circuit appear in the figures below and lead to the expressions given in the text.

Q.23. A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in Fig. 6.11. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field 

(i) Write down equation for the acceleration of the wire XY.

(ii) If B is independent of time, obtain v(t) , assuming v (0) = u₀.
(iii) For (b), show that the decrease in kinetic energy of XY equals the heat lost in R.

Ans: Consider the rod at position x(t) and moving with speed v. Magnetic flux through the rectangular loop is Φ = B · area = B · (x · ℓ) (area vector and B both along z).

Emf induced by changing flux: ε_flux = -dΦ/dt = -Bℓ·v. Motional emf across XY (if any) adds similarly; combined net emf around the loop is ε = -Bℓ·v (sign chosen by orientation). Current around the circuit is i = ε/R = -Bℓ v / R. Magnetic force on XY is F = iℓB (direction opposing motion).

Applying Newton's second law: m dv/dt = -(B²ℓ²/R) v. This is a first-order linear ODE of the form dv/dt = -(k/m) v with k = B²ℓ²/R. Solving with v(0) = u₀ gives v(t) = u₀ exp[-(B²ℓ²/mR) t].

Power dissipated as heat in R is P = i²R = (B²ℓ²/R) v². The rate of loss of kinetic energy is d(½mv²)/dt = m v dv/dt = -(B²ℓ²/R) v² which equals -P. Integrating over time shows the decrease in kinetic energy equals the energy dissipated as heat in R. The intermediate expressions and diagrams are given below.

Q.24. ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with an angular velocity ω (Fig 6.12). The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of λ per unit length. Find the current in the rotating conductor, as it rotates by 180°.

Ans: Key concept: As OP rotates, the contact length with the fixed conductor and the area of the loop change with time, producing a changing flux and hence an induced emf. Treat the instantaneous contact length OQ = x(t) and compute flux through the appropriate polygonal area. Apply Faraday's law to find the induced emf ε = -dΦ/dt, then use the resistance of the rotating conductor (proportional to contact length through λ) to get the induced current i = ε/R(x). Carry out this analysis in the three successive contact regimes (rod touching BD, then AB, then AC) as illustrated. The figures below show the geometry, flux expressions, resulting emf and the current formulas for each interval; substitution of x(t) = ωt·(appropriate factor) and the resistance per unit length λ gives the required time-dependent currents.

Q.25. Consider an infinitely long wire carrying a current I (t), with 

constant. 
Find the current produced in the rectangular loop of wire ABCD if its resistance is R (Fig. 6.13).

Ans: To find the flux through the rectangular loop, integrate the magnetic field due to the long straight wire over the area of the loop. For an element at distance r from the wire, B = μ₀I(t)/(2πr). An elementary area strip of width dr and length ℓ has dA = ℓ dr, so the flux Φ = ∫ B dA = (μ₀I(t)ℓ/2π) ∫(dr/r) between the appropriate limits. Differentiating the flux with respect to time gives ε = -dΦ/dt = -(μ₀ℓ/2πR) (dI/dt) ln(r₂/r₁). The induced current is i = ε/R and its sign is set by Lenz's law. The detailed integrals and algebra are shown in the figures below.

Q.26. A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I (t) = I₀ (l - t/T) for 0 < t < T and I(0) = 0 for t > T (Fig. 6.14). Find the total charge passing through a given point in the loop, in time T. The resistance of the loop is R.

Ans: The total charge Q that passes through a point is Q = ∫ i dt = (1/R) ∫ ε dt = (1/R) ∫ (-dΦ/dt) dt = -(1/R) [Φ(T) - Φ(0)]. Thus the total charge transferred during 0 to T equals the net change of magnetic flux through the loop divided by R (with sign according to orientation). Using the explicit I(t) and the expression for flux due to the nearby long wire (integral over 1/r), substitute I(t) into Φ and evaluate Φ(T) - Φ(0). The algebraic steps, integrals and simplifications are shown in the figures and lead to the expression given in the text for the total charge magnitude.

Q.27. A magnetic field B is confined to a region r ≤a and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge = Q) of radius b, b > a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time ∆t. Find the angular velocity ω of the ring after the field vanishes.

Ans: When B is reduced to zero, the flux through the area r ≤ a decreases. By Faraday's law an emf is induced around the ring, producing an electric field E and hence a tangential force on the charge Q. The torque τ = Q·E·b acts for time Δt producing an angular impulse equal to the change in angular momentum. Using ε = -dΦ/dt = ∮ E·dl and Φ = B₀πa² (initial flux), one finds E = (1/2πb)·(-ΔΦ/Δt) and τ = b·Q·E. Equating τΔt to the final angular momentum Iω (I = m b²) and simplifying gives ω = (Q Φ)/(2π m b²) where Φ is the initial flux (the figures show the detailed substitutions). The intermediate expressions and algebra are displayed below.

Q.28. A rod of mass m  and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ with respect to the horizontal (Fig. 6.15). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

Ans: Resolve B into components relative to the inclined plane: the component perpendicular to the plane (which produces motional emf) is B cos θ. The motional emf across the rod of length d moving at speed v is ε = v (B cos θ) d. The induced current is i = ε/R = vB d cos θ / R. Magnetic force on the rod is F_m = i d B sin 90° = (B² d² cos θ / R) v, directed up the incline and opposing motion. The weight component down the plane is mg sin θ. Newton's second law gives m dv/dt = mg sin θ - (B²d² cos θ / R) v. This is a first-order linear ODE with solution v(t) = (mg sin θ / γ)[1 - exp(-γt/m)] + u·exp(-γt/m) where γ = B²d² cos θ / R and u is the initial speed (u = 0 here). The explicit integrated steps and constants of integration are shown in the figures.

Q.29. Find the current in the sliding rod AB (resistance = R) for the arrangement shown in Fig 6.16. B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0.

Ans: The rod of length d moving at constant speed v in a uniform B⊥ produces a motional emf ε = vBd across its ends. With switch S closed, a capacitor (if present) charges to that emf; if there is no capacitor the induced emf drives current immediately. For a circuit containing R and a capacitor C, the current decays as the capacitor charges; the time dependence follows from i = C dε/dt + ε/R depending on circuit details. The specific solution for i(t) is obtained by writing KVL including ε(t) = vBd and solving the corresponding ODE; the stepwise integration and final expression for i(t) are given in the figures below.

Q.30. Find the current in the sliding rod AB (resistance = R) for the arrangement shown in Fig 6.17. B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0.

Ans: With the inductor present, the motional emf ε = vBd acts as the driving emf in the RL circuit. Writing KVL: L(di/dt) + iR = ε = vBd. This linear ODE with initial condition i(0) = 0 has solution i(t) = (vBd/R)[1 - exp(-R t/L)]. At long times the current approaches the steady value vBd/R. The derivation and integration steps are shown in the figures below.

Q.31. A metallic ring of mass m and radius l (ring being horizontal) is falling  under gravity in a region having a magnetic field. If z is the vertical direction, the z-component of magnetic field is B_z = B_o (1+λ z). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m , B, λ and acceleration due to gravity g.

Ans: Flux through the ring Φ = πl²B₀(1 + λ z). As the ring falls with speed v = dz/dt, the rate of change of flux is dΦ/dt = πl²B₀λ v. Induced emf ε = -dΦ/dt = -πl²B₀λ v. Induced current i = ε/R = -(πl²B₀λ/R) v. Power dissipated as heat is P = i²R = (π² l⁴ B₀² λ²/R) v². If the ring reaches terminal (constant) velocity v, mechanical power lost per second mgv equals electrical power dissipated: mgv = P. Solving for v gives v = (m g R) / (π² l⁴ B₀² λ²). The intermediate algebra and steps are provided in the figures below.

Q.32. A long solenoid 'S' has 'n' turns per meter, with diameter 'a'. At the centre of this coil we place a smaller coil of 'N' turns and diameter 'b' (where b < a). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil.

Plot graph showing nature of variation in emf, if current varies as a function of mt² + C.

Ans: Magnetic field inside a long solenoid is B = μ₀ n i(t). Flux through the small coil is Φ = N·B·A = N·μ₀ n i(t)·π b². Induced emf in the small coil is ε = -dΦ/dt = -N μ₀ n π b² di/dt. If i(t) increases linearly with time, di/dt is constant and ε is constant in magnitude (a rectangular pulse while di/dt remains constant). If i(t) = m t² + C, then di/dt = 2m t and ε(t) = -2m N μ₀ n π b² t, which varies linearly with time; the sign indicates opposition to the change. The graph of ε versus t is a straight line through the origin with slope -2m N μ₀ n π b² (the figures below illustrate the time dependence and sign convention).