CBSE Class 12 Physics Electromagnetic Induction Basic Problems

CBSE Class 12 Physics Electromagnetic Induction Basic Problems. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

BASIC PROBLEMS

1. A copper rod of length 0.19 m is moving with a uniform velocity 10 m/s parallel to a long straight wire carrying a current of 5.0 A. The rod itself is perpendicular to the wire with its ends at distances 0.01 m and 0.2 m from it. Calculate the e.m.f. induced in the rod. [3 Marks]

Answer:
1. Magnetic field at a distance x from a long straight current-carrying wire is given by B = \(\frac{\mu_0 I}{2\pi x}\).
2. When a small element of length dx is at distance x, the induced e.m.f. across dx moving with velocity v is d(e) = B v dx = \(\frac{\mu_0 I v}{2\pi x}\) dx.
3. Integrating from inner radius \(r_1 = 0.01\) m to outer radius \(r_2 = r_1 + l = 0.01 + 0.19 = 0.2\) m, total e.m.f. is E = \(\frac{\mu_0 I v}{2\pi} \ln\left(\frac{r_2}{r_1}\right)\).
4. Substituting the values: E = \(\frac{4\pi \times 10^{-7} \times 5 \times 10}{2\pi} \ln\left(\frac{0.2}{0.01}\right)\) = \(10^{-5} \ln(20)\) = \(10^{-5} \times 3.0\) = \(3.0 \times 10^{-5}\) V.

Teacher's Note:
a) The magnetic field due to a straight wire is non-uniform, requiring integration over the length of the rod.
b) Students often make errors in determining the correct limits of integration for the positions of the rod's ends.
c) Logarithm base conversion from natural to common log (\(\ln(x) = 2.303 \log_{10}(x)\)) is essential for final evaluation.

2. A long solenoid of diameter 0.1 m has \(2 \times 10^4\) turns per metre. At the centre of the solenoid a 100 turn coil of radius 0.01 m and resistance \(10\pi^2\) ohm is placed with its axis coinciding with the solenoid axis. The current in the solenoid is decreased at a constant rate from + 2 A to - 2 A in 0.05 s. Find the e.m.f. induced in the coil. Also find the total charge flowing through the coil during this time. [4 Marks]

Answer:
1. Magnetic field inside the solenoid is B = \(\mu_0 n I\).
2. Change in magnetic field \(\Delta B = \mu_0 n \Delta I = 4\pi \times 10^{-7} \times 2 \times 10^4 \times (-2 - 2) = -32\pi \times 10^{-3}\) T.
3. Induced e.m.f. is E = \(-N_2 \frac{\Delta \Phi}{\Delta t} = -N_2 A \frac{\Delta B}{\Delta t}\), where \(A = \pi r^2 = \pi (0.01)^2 = 10^{-4}\pi\) m\(^2\).
4. Substituting values: E = \(-100 \times 10^{-4}\pi \times \frac{-32\pi \times 10^{-3}}{0.05} = 6.4\pi^2 \times 10^{-3} \approx 0.063\) V (magnitude).
5. Total charge flowing is Q = \(\frac{E \Delta t}{R} = \frac{N_2 A \Delta B}{R} = \frac{100 \times 10^{-4}\pi \times 32\pi \times 10^{-3}}{10\pi^2} = 3.2 \times 10^{-5}\) C (= 32 \(\mu\)C).

Teacher's Note:
a) Faraday's law of electromagnetic induction is applied using the changing magnetic flux linked with the coil.
b) Charge flowing is independent of time and depends only on total flux change and resistance.
c) Check unit conversions for dimensions and maintain standard SI units throughout.
d) The negative sign indicates the direction of induced e.m.f. as per Lenz's law.

3. A very small circular loop of area \(5 \times 10^{-4}\) m\(^2\), resistance 2 ohm and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius 0.1 m. A constant current of 1 ampere is passed in the bigger loop and the smaller loop is rotated with angular velocity \(\omega\) rad/s about a diameter. Calculate as a function of time
(a) the flux linked with the smaller loop,
(b) induced e.m.f.
(c) induced current in the smaller loop. [4 Marks]

Answer:
1. Magnetic field at the centre of the large circular loop is B = \(\frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 1}{2 \times 0.1} = 2\pi \times 10^{-6}\) T.
2. (a) Magnetic flux linked with the smaller loop as a function of time is \(\Phi(t) = B A \cos(\omega t) = 2\pi \times 10^{-6} \times 5 \times 10^{-4} \cos(\omega t) = \pi \times 10^{-9} \cos(\omega t)\) Wb.
3. (b) Induced e.m.f. is E = \(-\frac{d\Phi}{dt} = \omega \pi \times 10^{-9} \sin(\omega t)\) V.
4. (c) Induced current in the smaller loop is I_ind = \(\frac{E}{R} = \frac{\omega \pi \times 10^{-9} \sin(\omega t)}{2} = 0.5 \omega \pi \times 10^{-9} \sin(\omega t)\) A.

Teacher's Note:
a) Magnetic field is assumed uniform over the small loop area because its dimensions are much smaller than the outer loop radius.
b) Time variation of flux arises purely due to the changing orientation angle \(\theta = \omega t\) of the rotating loop.
c) Ohm's law is applied directly to find the induced current from the induced e.m.f. and resistance.
d) Trigonometric derivative rules are utilized correctly to differentiate cosine into sine for e.m.f.

4. A closed coil consists of 500 turns on a a rectangular frame of area \(4.0\) cm\(^2\) and has a resistance of 50 ohm. The coil is kept with its plane perpendicular to a uniform magnetic field of 0.2 T. Calculate the amount of charge flowing through the coil if it is turned over (rotated through 180°). Will the answer depend on the speed with which the coil is rotated ? [3 Marks]

Answer:
1. Initial flux linked with the coil is \(\Phi_1 = N B A \cos(0^\circ) = 500 \times 0.2 \times 4.0 \times 10^{-4} \times 1 = 0.04\) Wb.
2. Final flux after 180° rotation is \(\Phi_2 = N B A \cos(180^\circ) = -0.04\) Wb.
3. Total charge flowing is Q = \(\frac{\Delta \Phi}{R} = \frac{\Phi_1 - \Phi_2}{R} = \frac{0.04 - (-0.04)}{50} = \frac{0.08}{50} = 1.6 \times 10^{-3}\) C.
4. No, the amount of charge flowing does not depend on the speed of rotation because charge depends only on total flux change and resistance.

Teacher's Note:
a) Total charge formula Q = \(\Delta\Phi / R\) is derived by integrating current over time during the rotation interval.
b) Rotation through 180° reverses the sign of flux contribution, doubling the effective flux change.
c) Speed affects the magnitude of induced current and peak e.m.f., but not the total integrated charge.

5. A long solenoid having 1000 turns per cm carries an alternating current of peak value 1 ampere. A search coil having a cross-sectional area of \(1 \times 10^{-4}\) m\(^2\) and 20 turns is kept in the solenoid so that its plane is perpendicular to the axis of the solenoid. The search coil registers a peak voltage of \(2.5 \times 10^{-2}\) volt. Find the frequency of the current in the solenoid. [3 Marks]

Answer:
1. Let the alternating current be \(I = I_0 \sin(\omega t)\), where \(I_0 = 1\) A.
2. Magnetic field inside the solenoid is B = \(\mu_0 n I_0 \sin(\omega t)\), where n = \(1000 \text{ turns/cm} = 10^5 \text{ turns/m}\).
3. Flux linked with the search coil of N turns is \(\Phi = N_s B A = N_s (\mu_0 n I_0 \sin(\omega t)) A\).
4. Induced peak e.m.f. is E₀ = \(N_s \mu_0 n I_0 A \omega\), which gives \(2.5 \times 10^{-2} = 20 \times (4\pi \times 10^{-7}) \times 10^5 \times 1 \times (1 \times 10^{-4}) \times (2\pi f)\).
5. Solving for frequency f: f = \(\frac{2.5 \times 10^{-2}}{16\pi^2 \times 10^{-5}} \approx 15.8\) Hz.

Teacher's Note:
a) Turn density n must be converted properly from turns per cm to turns per meter.
b) Peak voltage is equated to the maximum derivative of magnetic flux with respect to time.
c) Angular frequency is related to ordinary frequency by \(\omega = 2\pi f\).

6. A coil of inductance 2 H and resistance 10 ohm is connected across a battery of 100 V along with a key. At time t = 0, the key is closed. Find
(a) time constant of the circuit
(b) steady-state current in the circuit
(c) current in the circuit at t = 1 s
(d) magnetic energy stored in the coil in the steady state. [4 Marks]

Answer:
1. (a) Time constant \(\tau = \frac{L}{R} = \frac{2}{10} = 0.2\) s.
2. (b) Steady-state current \(I_0 = \frac{E}{R} = \frac{100}{10} = 10\) A.
3. (c) Current at time t is given by \(I = I_0 (1 - e^{-t/\tau})\). Substituting values for t = 1 s: \(I = 10 (1 - e^{-1/0.2}) = 10 (1 - e^{-5}) = 10 (1 - 0.0067) = 9.933\) A.
4. (d) Magnetic energy stored in steady state is U = \(\frac{1}{2} L I_0^2 = \frac{1}{2} \times 2 \times (10)^2 = 100\) J.

Teacher's Note:
a) Growth of current in an L-R circuit follows an exponential formula governed by the circuit time constant.
b) In the steady state, the inductor acts like a simple conducting wire with zero potential drop across it.
c) Exponential value \(e^{-5} \approx 0.0067\) must be computed or substituted accurately.
d) Magnetic energy formula depends directly on inductance and square of steady-state current.

7. Two long parallel wires of zero resistance are connected to each other by a battery of 1.0 volt. The separation between the wires is 0.5 m. A metallic bar, which is perpendicular to the wires and of resistance 10 ohm moves on these wires while a magnetic field of 0.02 Tesla is acting perpendicular to the plane containing the bar and the wires. Find the steady state velocity of the bar. If mass of the bar is 0.002 kg, find its velocity as a function of time. [4 Marks]

Answer:
1. When the bar moves with velocity v, back e.m.f. induced is E_ind = B l v.
2. Net current through the bar is I = \(\frac{V - B l v}{R}\).
3. Magnetic force acting on the bar is \(F_m = I l B = \frac{(V - B l v)l B}{R}\). In steady state, acceleration is zero, so \(F_m = 0\), giving steady-state velocity v_steady = \(\frac{V}{B l} = \frac{1.0}{0.02 \times 0.5} = 100\) m/s.
4. Equation of motion for mass m is \(m \frac{dv}{dt} = \frac{(V - B l v)l B}{R}\), leading to \(\frac{dv}{dt} + \frac{B^2 l^2}{m R} v = \frac{B l V}{m R}\). Solving this differential equation gives velocity as a function of time: \(v(t) = v_{\text{steady}} (1 - e^{-t/\tau_m})\), where time constant \(\tau_m = \frac{m R}{B^2 l^2} = \frac{0.002 \times 10}{(0.02)^2 \times (0.5)^2} = \frac{0.02}{0.0001} = 200\) s, hence \(v(t) = 100(1 - e^{-t/200})\) m/s.

For \(t \ll 200\) s the bar starts with acceleration \(\frac{V l B}{m R} = 0.5\) m/s\(^2\), so \(v \approx 0.5t\) m/s, which is the answer given in the key.

Teacher's Note:
a) Motion of the conducting bar creates a motional e.m.f. that opposes the battery voltage.
b) Steady-state condition occurs when the magnetic force balances out the driving electromagnetic forces to yield zero net force.
c) Newton's second law combined with electromagnetic induction sets up a standard first-order differential equation for velocity.
d) Mechanical time constant depends on mass, resistance, magnetic field, and separation distance.

8. The network shown in Fig. is a part of a complete circuit. What is the potential difference \(V_B - V_A\), when the current I is 5 A and is decreasing at a rate of \(10^3\) (A/s). [3 Marks]

[Figure: A network segment from A to B showing a resistor of 1 ohm, a battery of 15 V, and an inductor of 5 mH connected in series]

Answer:
1. The direction of current is from A to B, with current I = 5 A decreasing at the rate \(\frac{dI}{dt} = -10^3\) A/s.
2. Applying Kirchhoff's loop rule starting from point A to point B: \(V_A - I(1) + 15 - L\left(\frac{dI}{dt}\right) = V_B\).
3. Substituting the given values: \(V_A - (5)(1) + 15 - (5 \times 10^{-3})(-10^3) = V_B\).
4. Simplifying the equation: \(V_A - 5 + 15 + 5 = V_B \implies V_B = V_A + 15\).
5. Therefore, the potential difference \(V_B - V_A = 15\) V.

Teacher's Note:
a) The induced emf across the inductor opposes the change in current, given by \(e = -L\frac{dI}{dt}\).
b) Pay close attention to the polarity and sign conventions when traversing against or along the decreasing current.
c) Convert millihenry to henry correctly by multiplying by \(10^{-3}\).

9. The figure shows a horizontal conductor AC of mass 0.1 kg and length 1 m sliding along two smooth parallel conducting rails PQ and ST, inclined at 30° to the horizontal. At the top, the rails are connected by a resistance R = 2 \(\Omega\). The system is in a vertical uniform magnetic field B = 1.0 T. What is the terminal speed of the conductor AC and the current in the loop APSC when AC has achieved terminal speed? Neglect the resistance of rest of the circuit. [5 Marks]

Answer:
1. The component of gravitational force acting down the inclined plane is \(F_g = mg \sin\theta\).
2. When the conductor moves with velocity v, the motional emf induced is \(e = Blv \cos\theta\) (since the magnetic field is vertical and the component of B perpendicular to the incline is \(B \cos\theta\)).
3. The induced current in the loop is \(I = \frac{e}{R} = \frac{Blv \cos\theta}{R}\).
4. The magnetic force acting up the incline is \(F_m = IlB \cos\theta = \frac{B^2 l^2 v \cos^2\theta}{R}\).
5. At terminal speed, the component of gravitational force down the incline equals the magnetic force up the incline: \(mg \sin\theta = \frac{B^2 l^2 v \cos^2\theta}{R}\).
6. Solving for terminal speed v: \(v = \frac{mg R \sin\theta}{B^2 l^2 \cos^2\theta}\).
7. Substituting the values (\(m = 0.1\), \(g = 9.8\) or \(10\), \(\theta = 30^\circ\), \(R = 2\), \(B = 1.0\), \(l = 1\)): \(v = \frac{0.1 \times 9.8 \times 2 \times \sin(30^\circ)}{(1.0)^2 (1)^2 \cos^2(30^\circ)} = \frac{1.96 \times 0.5}{1 \times (\frac{\sqrt{3}}{2})^2} = \frac{0.98}{\frac{3}{4}} = 1.31\) m/s (using g = 9.8 m/s²).
8. The current at terminal speed is \(I = \frac{Blv \cos\theta}{R} = \frac{1.0 \times 1 \times 1.31 \times \cos(30^\circ)}{2} = 0.57\) A.

Teacher's Note:
a) The effective perpendicular component of the magnetic field to the inclined plane must be carefully resolved using the angle of inclination.
b) A common error is neglecting the angle in the motional emf and magnetic force formulas.
c) Terminal velocity is achieved when net acceleration becomes zero, balancing gravity and electromagnetic damping.
d) Always state units clearly at each final numerical evaluation step.

10. A rectangular frame ABCD made of a uniform metal wire has a straight connection between E and F made of the same wire, as shown in the figure. AEFD is a square of side 1 m and EB = FC = 0.5 m. The entire circuit is placed in a steadily increasing, uniform magnetic field directed into the plane of the paper and normal to it. The rate of change of magnetic field is 1 T/s. The resistance per unit length of the wire is 1 \(\Omega\)/m. Find the magnitudes and directions of the currents in the segments AE, BE and EF. [5 Marks]

[Figure: Rectangular frame ABCD with internal wire EF connecting side AB and CD, forming square AEFD of side 1 m and rectangular region EBCF of width 0.5 m]

Answer:
1. Given side of square AEFD = 1 m, so length AE = DF = EF = AD = 1 m. Also EB = FC = 0.5 m, AB = CD = 1.5 m.
2. Resistance per unit length \(\lambda = 1\).
3. Resistance of segments: \(R_{AE} = 1 \times 1 = 1 \, \Omega\), \(R_{ED} = 1 \times 1 = 1 \, \Omega\), \(R_{DF} = 1 \times 1 = 1 \, \Omega\), \(R_{FA} = 1 \times 1 = 1 \, \Omega\), \(R_{EF} = 1 \times 1 = 1 \, \Omega\), \(R_{BE} = 1 \times 0.5 = 0.5 \, \Omega\), \(R_{CF} = 1 \times 0.5 = 0.5 \, \Omega\), \(R_{BC} = 1 \times 1 = 1 \, \Omega\).
4. Induced emf in square loop AEFD is \(e_1 = \left(\frac{dB}{dt}\right) \times \text{Area}_1 = 1 \times (1 \times 1) = 1\) V.
5. Induced emf in loop EBCF is \(e_2 = \left(\frac{dB}{dt}\right) \times \text{Area}_2 = 1 \times (0.5 \times 1) = 0.5\) V.
6. Applying mesh analysis or Kirchhoff's laws to the two loops with anticlockwise induced currents due to inward increasing magnetic field: let current in loop AEFD be \(I_1\) and in loop EBCF be \(I_2\).
7. Writing mesh equations for the counter-clockwise currents \(I_1\) (loop AEFD, resistance 4 \(\Omega\)) and \(I_2\) (loop EBCF, resistance 3 \(\Omega\)) with EF shared (1 \(\Omega\)): \(1 = 4I_1 - I_2\) and \(0.5 = 3I_2 - I_1\). Hence \(I_1 = \frac{7}{22}\) A and \(I_2 = \frac{3}{11}\) A. Current in AE = \(\frac{7}{22}\) A, directed from E to A; current in BE = \(\frac{3}{11}\) A, directed from B to E; current in EF = \(I_1 - I_2 = \frac{1}{22}\) A, directed from F to E.

Teacher's Note:
a) Divide the complex network into individual adjacent loops to apply Faraday's law and mesh currents systematically.
b) Miscalculating individual wire lengths and their corresponding resistances is the most frequent student error.
c) Lenz's law dictates that the induced current creates an outward magnetic field to oppose the inward increasing magnetic field.
d) Double-check loop boundaries and shared branch current orientations before finalizing mesh equations.

11. A current of 10 A is flowing in a long straight wire situated near a rectangular circuit whose two sides of length 0.2 m are parallel to the wire. One of them is at a distance of 0.05 m and the other is at a distance of 0.10 m from the long wire. The wire is in the plane of the rectangle. Find the magnetic flux through the rectangular circuit. If the current decays uniformly to zero in 0.02 s, find the e.m.f induced in the circuit. [5 Marks]

[Figure: A long straight wire carrying current 10 A next to a rectangular loop of dimensions 0.2 m by (0.10 - 0.05) m, with distances from wire marked as 0.05 m and 0.10 m]

Answer:
1. Magnetic field at a distance r from a long straight wire carrying current I is \(B = \frac{\mu_0 I}{2\pi r}\).
2. Consider a strip of width dr at distance r from the wire, parallel to the wire, within the rectangular loop of length l = 0.2 m.
3. Total magnetic flux \(\Phi\) through the rectangular circuit is given by \(\Phi = \int_{r_1}^{r_2} B l \, dr = \frac{\mu_0 I l}{2\pi} \int_{r_1}^{r_2} \frac{dr}{r} = \frac{\mu_0 I l}{2\pi} \ln\left(\frac{r_2}{r_1}\right)\).
4. Substituting the given values (\(I = 10\) A, \(l = 0.2\) m, \(r_1 = 0.05\) m, \(r_2 = 0.10\) m, \(\mu_0 = 4\pi \times 10^{-7}\)):
\(\Phi = \frac{(4\pi \times 10^{-7})(10)(0.2)}{2\pi} \ln\left(\frac{0.10}{0.05}\right)\)
5. \(\Phi = 4 \times 10^{-7} \ln(2) = 4 \times 10^{-7} \times 0.693 = 2.77 \times 10^{-7}\) Wb.
6. Induced emf \(e = \frac{d\Phi}{dt} = \frac{\Phi_{\text{initial}} - 0}{\Delta t} = \frac{2.77 \times 10^{-7}}{0.02} = 1.39 \times 10^{-5}\) V.

Teacher's Note:
a) Integration is essential here because the magnetic field varies inversely with distance from the straight wire.
b) Ensure proper logarithmic conversion from natural log to base-10 if required, or keep it as natural logarithm \(\ln(2) \approx 0.693\).
c) The rate of change of flux directly yields the magnitude of the induced emf according to Faraday's law.
d) Verify distance limits \(r_1\) and \(r_2\) correctly corresponding to the inner and outer edges of the rectangle.

12. A circular coil P of 100 turns and radius 2 cm is placed coaxially at the centre of another circular coil Q of 100 turns and radius 20 cm. Calculate
(a) the mutual inductance of the coils,
(b) the induced e.m.f. in coil P when the current in the coil Q decreases from 5 amp. to 3 amp.
(c) the rate of change of flux through the coil P at this instant and
(d) the charge passing through coil P if its resistance is 8 ohm. [5 Marks]

Answer:
1. (a) Mutual inductance M between two concentric coplanar coils is given by \(M = \frac{\mu_0 \pi N_1 N_2 r_1^2}{2 R_2}\), where \(r_1 = 0.02\) m, \(R_2 = 0.20\) m, \(N_1 = 100\), \(N_2 = 100\).
\(M = \frac{(4\pi \times 10^{-7}) \pi (100)(100)(0.02)^2}{2(0.20)} = \frac{4\pi^2 \times 10^{-3} \times 0.0004}{0.40} = 3.95 \times 10^{-5}\) H.
2. (b) Induced emf in coil P is \(e = M \frac{dI}{dt}\). Average induced emf for current change \(\Delta I = 5 - 3 = 2\) A in time interval \(\Delta t\) is needed, or using general formula. Assuming a standard time or calculating flux linkage change: since time dt is not specified, part (b) evaluates \(e = M \frac{\Delta I}{\Delta t}\) or is expressed in terms of \(\Delta t\). Alternatively, if interval is implied or calculated via charge: \(Q_{\text{charge}} = \frac{\Delta \Phi}{R} = \frac{M \Delta I}{R} = \frac{(3.95 \times 10^{-5})(2)}{8} = 9.87 \times 10^{-6}\) C.
3. (c) Rate of change of flux through coil P equals induced emf magnitude \(e = \frac{d\Phi_P}{dt}\).
4. (d) Total charge passing \(q = \frac{\Delta \Phi_{\text{total}}}{R} = \frac{M \Delta I}{R} = 9.87 \times 10^{-6}\) C.

Answer key values: M = 3.94 × 10^-4 H, induced e.m.f. = 1.97 × 10² V, rate of change of flux = 1.97 × 10^-2 Wb/s, charge = 9.86 × 10^-5 C. (With the data as printed, N = 100 for both coils, the formula gives M = 3.95 × 10^-5 H; the key values correspond to a mutual inductance ten times larger. The time of the current change is not given in the problem.)

Teacher's Note:
a) The mutual inductance formula for a small central coil inside a large outer coil relies on a uniform magnetic field approximation over the inner coil's area.
b) Charge flow is independent of time and depends only on total flux linkage change and resistance.
c) Check unit conversions carefully for radii from centimeters to meters.
d) Rate of change of flux and induced emf are numerically equal in magnitude.

13. A solenoid of resistance 50 \(\Omega\) and inductance 80 henry are connected to a 200 volt battery. How long will it take for the current to reach 50% of its final equilibrium value ? Calculate the maximum energy stored. [3 Marks]

Answer:
1. Final equilibrium current \(I_0 = \frac{V}{R} = \frac{200}{50} = 4\) A.
2. Time constant of the LR circuit is \(\tau = \frac{L}{R} = \frac{80}{50} = 1.6\) s.
3. The growth of current in an LR circuit is given by \(I = I_0 (1 - e^{-t/\tau})\).
4. Given \(I = 50\% \text{ of } I_0 = 0.5 I_0\), so \(0.5 I_0 = I_0 (1 - e^{-t/\tau}) \implies e^{-t/\tau} = 0.5\).
5. Taking natural logarithm: \(\frac{t}{\tau} = \ln(2) = 0.693 \implies t = 1.6 \times 0.693 = 1.11\) s.
6. Maximum energy stored in the inductor is \(U_{\text{max}} = \frac{1}{2} L I_0^2 = \frac{1}{2} (80)(4)^2 = 40 \times 16 = 640\) J.

Teacher's Note:
a) The growth of current formula \(I = I_0(1 - e^{-t/\tau})\) governs LR circuit transient behavior.
b) Half-life equivalent time for current growth is always \(t = \tau \ln(2)\).
c) Maximum energy is stored when the current reaches its steady maximum equilibrium value.

14. A copper rod of length 0.19 m is moving with uniform velocity 10 m/s parallel to a long straight wire carrying a current of 5.0 ampere. The rod itself is perpendicular to the wire with its ends at distances 0.01 m and 0.2 m from it. Calculate the emf induced in the rod. [3 Marks]

Answer:
1. Magnetic field at distance r from a long straight wire carrying current I is \(B = \frac{\mu_0 I}{2\pi r}\).
2. The rod moves with velocity v parallel to the wire, cutting magnetic field lines. The induced emf across an infinitesimal element dr at distance r is \(de = B v \, dr\).
3. Total induced emf is obtained by integrating from inner distance \(r_1 = 0.01\) m to outer distance \(r_2 = 0.2\) m:
\(e = \int_{r_1}^{r_2} \frac{\mu_0 I v}{2\pi r} dr = \frac{\mu_0 I v}{2\pi} \ln\left(\frac{r_2}{r_1}\right)\).
4. Substituting the given values (\(I = 5.0\) A, \(v = 10\) m/s, \(r_1 = 0.01\) m, \(r_2 = 0.2\) m, \(\mu_0 = 4\pi \times 10^{-7}\)):
\(e = \frac{(4\pi \times 10^{-7})(5.0)(10)}{2\pi} \ln\left(\frac{0.2}{0.01}\right)\)
5. \(e = 10^{-5} \ln(20) = 10^{-5} \times 2.996 \approx 3.0 \times 10^{-5}\) V = 30 \(\mu\)V.

Teacher's Note:
a) Integration is required because the magnetic field intensity decreases with distance from the straight wire.
b) Take care to use natural logarithm \(\ln(20) \approx 2.996\) correctly.
c) The motion is parallel to the wire, so motional emf is generated across the length spanning the two radial distances.

15. An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 \(\mu\)F and the resulting L-C circuit is set oscillating at its natural frequency. Let q denote the instantaneous charge on the capacitor, and I the current in the circuit. It is found that the maximum value of q is 200 \(\mu\)C. [5 Marks]
(a) When q = 100 \(\mu\)C what is the value of |dI/dt|?
(b) When q = 200 \(\mu\)C, what is the value of I?
(c) Find the maximum value of I?
(d) When I is equal to one half its maximum value what is the value of |q|?

Answer:
1. Given parameters are L = 2.0 mH = 2 × 10⁻³ H, C = 5.0 \(\mu\)F = 5 × 10⁻⁶ F, and maximum charge \(q_{max}\) = 200 \(\mu\)C = 2 × 10⁻⁴ C.
2. Angular frequency \(\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 10^{-3} \times 5 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-8}}} = 10^4\) rad/s.
3. (a) For part (a), using Kirchhoff's voltage law or the loop equation, \(L \frac{dI}{dt} + \frac{q}{C} = 0\), hence \(\left|\frac{dI}{dt}\right| = \frac{q}{LC}\). Substituting q = 100 \(\mu\)C = \(10^{-4}\) C gives \(\left|\frac{dI}{dt}\right| = \frac{10^{-4}}{2 \times 10^{-3} \times 5 \times 10^{-6}} = 10^4\) A/s.
4. (b) When q = 200 \(\mu\)C = \(q_{max}\), the capacitor has maximum charge and instantaneous current I = 0.
5. (c) The maximum current is \(I_{max} = q_{max}\omega = 2 \times 10^{-4} \times 10^4 = 2\) A.
6. (d) Using the conservation of energy equation for LC oscillations, \(I^2 = \omega^2 (q_{max}^2 - q^2)\). When \(I = \frac{I_{max}}{2} = 1\) A, substituting \(I_{max} = q_{max}\omega\) yields \(\left(\frac{q_{max}}{2}\right)^2 = q_{max}^2 - q^2\), which gives \(q = \frac{\sqrt{3}}{2} q_{max} = \frac{\sqrt{3}}{2} \times 200\) \(\mu\)C = \(100\sqrt{3}\) \(\mu\)C.

Teacher's Note:
a) The angular frequency formula \(\omega = \frac{1}{\sqrt{LC}}\) is fundamental to analyzing LC circuit oscillations.
b) Students must remember that maximum current occurs when the charge on the capacitor is zero.
c) Energy conservation between electrostatic potential energy in the capacitor and magnetic energy in the inductor simplifies finding charge at intermediate currents.
d) Verify all unit conversions for capacitance from microfarads and inductance from millihenries.

16. A Square frame with side a and a long straight conductor carrying a constant current I are located in the same plane. The inductance and the resistance of the frame are equal to L and R respectively. The frame was turned through 180° about the axis OO′ separated from the current carrying conductor by a distance b. Find the electric charge having flown through the frame. [5 Marks]

Answer:
1. The magnetic field at a perpendicular distance x from a long straight wire carrying current I is given by \(B = \frac{\mu_0 I}{2\pi x}\).
2. Consider a strip of the square frame of thickness dx at a distance x from the wire, where area element is \(dA = a \, dx\).
3. The total initial magnetic flux linked with the frame is \(\Phi_1 = \int_{b-a/2}^{b+a/2} \frac{\mu_0 I a}{2\pi x} dx = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{2b+a}{2b-a}\right)\).
4. When the frame is turned through 180°, the direction of the normal vector relative to the magnetic field reverses, so the final flux is \(\Phi_2 = -\Phi_1\).
5. The total change in magnetic flux is \(\Delta\Phi = \Phi_2 - \Phi_1 = -\frac{\mu_0 I a}{\pi} \ln\left(\frac{2b+a}{2b-a}\right)\).
6. The total electric charge flown through the frame is \(q = \frac{|\Delta\Phi|}{R} = \frac{\mu_0 I a}{\pi R} \ln\left(\frac{2b+a}{2b-a}\right)\).

Teacher's Note:
a) Charge flown is given by the integral of current over time, which equals the total change in flux divided by resistance R.
b) A common error is missing the factor of 2 when the loop is rotated by 180 degrees.
c) Integration limits extend from b - a/2 to b + a/2 across the width of the square frame, since the axis OO' is at distance b from the wire. The key gives \(\frac{\mu_0 a I}{2\pi R}\ln\frac{a+b}{b-a}\).
d) Dimensionally, the logarithmic term is dimensionless, and the resulting expression yields units of Coulombs.

17. Two concentric coplanar circular loops are made of wire with resistance per unit length \(10^{-4}\) \(\Omega\) m\(^{-1}\), and have diameters 0.2 m and 2 m. A time varying potential difference (4 + 2.5 t) volts is applied to the larger loop. Calculate the current in the smaller loop. [5 Marks]

Answer:
1. Let the radius of the smaller loop be \(r_1 = \frac{0.2}{2} = 0.1\) m, and the radius of the larger loop be \(r_2 = \frac{2}{2} = 1.0\) m.
2. The resistance per unit length is \(\lambda = 10^{-4}\) \(\Omega\) m\(^{-1}\). The total resistance of the larger loop is \(R_2 = \lambda (2\pi r_2) = 10^{-4} \times 2\pi (1.0) = 2\pi \times 10^{-4}\) \(\Omega\).
3. The current in the larger loop as a function of time is \(I_2(t) = \frac{V(t)}{R_2} = \frac{4 + 2.5t}{2\pi \times 10^{-4}}\).
4. The magnetic field at the center of the larger loop produced by its current \(I_2\) is \(B = \frac{\mu_0 I_2}{2r_2}\). Assuming the smaller loop is at the center and experiences this uniform field across its area \(A_1 = \pi r_1^2\), the magnetic flux linked with the smaller loop is \(\Phi_1 = B A_1 = \left(\frac{\mu_0 I_2}{2r_2}\right) (\pi r_1^2)\).
5. Substituting values: \(\Phi_1 = \frac{4\pi \times 10^{-7} \times (4 + 2.5t)}{2(1.0) \times 2\pi \times 10^{-4}} \times \pi (0.1)^2 = \pi \times 10^{-3} (4 + 2.5t)\) Wb.
6. The induced emf in the smaller loop is \(e = -\frac{d\Phi_1}{dt} = -\frac{d}{dt} [\pi \times 10^{-3} (4 + 2.5t)] = -2.5\pi \times 10^{-3}\) V.
7. The resistance of the smaller loop is \(R_1 = \lambda (2\pi r_1) = 10^{-4} \times 2\pi (0.1) = 0.2\pi \times 10^{-4}\) \(\Omega\).
8. The current in the smaller loop is \(I_1 = \frac{|e|}{R_1} = \frac{2.5\pi \times 10^{-3}}{0.2\pi \times 10^{-4}} = \frac{2.5 \times 10^{-3}}{2 \times 10^{-5}} = 125\) A.

Teacher's Note:
a) Mutual inductance concept or direct field formulation from the circular loop center is used here.
b) Be careful to distinguish between the radii and resistances of the two distinct circular loops.
c) The time derivative of the potential difference yields a constant induced emf because the potential varies linearly with time.
d) Check all powers of 10 carefully during division of resistance and flux values.

18. The current (in ampere) in an inductor is given by I = 5 + 16t, where t is in second. The self-induced emf in it is 10 mV. Find: [3 Marks]
(a) the self-inductance, and
(b) the energy stored in the inductor and the power supplied to it at t = 1 sec.

Answer:
1. (a) The rate of change of current is \(\frac{dI}{dt} = \frac{d}{dt}(5 + 16t) = 16\) A/s.
2. The magnitude of self-induced emf is \(e = L \frac{dI}{dt}\), so \(10 \times 10^{-3} = L \times 16\), which gives self-inductance \(L = \frac{10^{-2}}{16} = 6.25 \times 10^{-4}\) H = 0.625 mH.
3. (b) At t = 1 s, the current is \(I = 5 + 16(1) = 21\) A. The energy stored in the inductor is \(U = \frac{1}{2} L I^2 = \frac{1}{2} \times 6.25 \times 10^{-4} \times (21)^2 = 0.5 \times 6.25 \times 10^{-4} \times 441 = 0.1378\) J.
4. The power supplied to the inductor at t = 1 s is \(P = e I = (10 \times 10^{-3}\text{ V}) \times (21\text{ A}) = 0.21\) W.

Teacher's Note:
a) Self-induced emf is calculated using the formula \(e = L \frac{dI}{dt}\).
b) Energy stored in an inductor is given by \(U = \frac{1}{2} L I^2\), and instantaneous power is \(P = e I\).
c) Always convert millivolts to volts and use standard SI units throughout calculations.

19. In figure there is a uniform magnetic field in the x-direction with B = 0.2 T. The loop has an area of 5 cm\(^2\) and rotates about line CD as axis. Point A rotates towards positive x values from the position shown. If AE rotates through 50° from its indicated position in a time of 0.2 s [5 Marks]
(a) What is the change in flux through the coil?
(b) What is the average induced emf in it?
(c) Does the induced current flow from A to C in the upper part of the coil?

[Figure: A rectangular loop oriented in a magnetic field directed along the x-axis, with axis of rotation CD, and points A and E indicated on the loop]

Answer:
1. (a) Initially the plane of the loop is perpendicular to the field (normal along the x-direction), so \(\theta_1 = 0^\circ\) and \(\Phi_1 = B A \cos(0^\circ) = 0.2 \times 5 \times 10^{-4} = 100\ \mu\)Wb.
2. When AE rotates through \(50^\circ\), the angle between the normal and the field becomes \(\theta_2 = 50^\circ\), so \(\Phi_2 = B A \cos(50^\circ) = 10^{-4} \times 0.643 \approx 64\ \mu\)Wb.
3. The change in flux is \(\Delta\Phi = \Phi_2 - \Phi_1 \approx 64 - 100 = -36\ \mu\)Wb.
4. (b) The average induced emf is \(\varepsilon = -\frac{\Delta\Phi}{\Delta t} = \frac{36 \times 10^{-6}}{0.2} = 180\ \mu\)V.
5. (c) Using Lenz's law, the flux decreases, so the induced current opposes the decrease and flows from A to C in the upper part of the coil: Yes.

Teacher's Note:
a) Magnetic flux is calculated as \(\Phi = B A \cos\theta\) where \(\theta\) is the angle between the magnetic field and the area vector.
b) Average induced emf is given by Faraday's law: magnitude of flux change divided by time interval.
c) Direction of induced current is determined by applying Lenz's law to oppose the flux change.
d) Ensure area is converted from square centimeters to square meters properly.

20. A capacitor discharges through an inductance of 0.1 henry and a resistance of 100 ohm. If the frequency of discharge is 1000 Hz, calculate the capacitance. [3 Marks]

Answer:
1. The frequency of damped oscillations in an LCR circuit (or discharging capacitor circuit) is given by \(f = \frac{1}{2\pi} \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}\).
2. Given values: L = 0.1 H, R = 100 \(\Omega\), and frequency f = 1000 Hz.
3. Squaring both sides: \((2\pi f)^2 = \frac{1}{LC} - \frac{R^2}{4L^2}\).
4. Calculate individual terms: \(2\pi f = 2 \times \pi \times 1000 = 2000\pi\) rad/s, so \((2\pi f)^2 = 4 \times 10^6 \pi^2 \approx 4 \times 10^6 \times 9.8696 = 3.948 \times 10^7\).
5. Calculate damping term: \(\frac{R^2}{4L^2} = \frac{(100)^2}{4(0.1)^2} = \frac{10000}{4 \times 0.01} = \frac{10000}{0.04} = 250,000 = 2.5 \times 10^5\).
6. Rearranging for capacitance C: \(\frac{1}{LC} = (2\pi f)^2 + \frac{R^2}{4L^2} = 3.948 \times 10^7 + 0.025 \times 10^7 = 3.973 \times 10^7\).
7. Therefore, \(C = \frac{1}{L(3.973 \times 10^7)} = \frac{1}{0.1 \times 3.973 \times 10^7} = \frac{1}{3.973 \times 10^6} \approx 2.52 \times 10^{-7}\) F = 0.252 \(\mu\)F.

Teacher's Note:
a) The frequency formula accounts for both the natural frequency term and the damping effect due to resistance.
b) Neglecting the resistance term when it is small is a common approximation, but here R is significant and must be included.
c) Verify arithmetic steps involving squares of pi and powers of ten carefully to avoid rounding errors.

SECTION A

1. Answer Key Item 1 [1 Mark]

Answer: \(3 \times 10^{-5}\) V

Teacher's Note:
a) This is an answer key entry corresponding to problem 1 of the exercise set.
b) Verify magnitude and powers of ten carefully during calculation.

2. Answer Key Item 2 [1 Mark]

Answer: 32 \(\mu\)C

Teacher's Note:
a) This value represents a charge computed from capacitance and potential difference.
b) Pay close attention to micro-coulomb units.

3. Answer Key Item 3 [3 Marks]

Answer:
(a) \(\pi \times 10^{-9} \cos \omega t\) weber
(b) \(\pi\omega \times 10^{-9} \sin \omega t\) volt
(c) \(\frac{\pi\omega}{2} \times 10^{-9} \sin \omega t\) ampere

Teacher's Note:
a) The three parts denote magnetic flux, induced emf, and induced current respectively.
b) Differentiation with respect to time yields the emf expression.
c) Ohm's law relates the induced current to the emf and resistance.

4. Answer Key Item 4 [1 Mark]

Answer: \(1.6 \times 10^{-3}\) C

Teacher's Note:
a) This is the final evaluated charge in coulombs.
b) Standard scientific notation is maintained.

5. Answer Key Item 5 [1 Mark]

Answer: 15.8 Hz

Teacher's Note:
a) Frequency is expressed in hertz.
b) Ensure correct conversion between angular frequency and linear frequency.

6. Answer Key Item 6 [3 Marks]

Answer: 0.2 S, 9.93 A, 100 J

Teacher's Note:
a) Multiple parameters involving conductance, current, and energy are listed.
b) Units must be verified for each individual physical quantity.
c) Energy is correctly stated in joules.

7. Answer Key Item 7 [2 Marks]

Answer: 100 m/s, 0.5t

Teacher's Note:
a) Represents velocity and a time-dependent expression.
b) Dimensional consistency must be maintained across variables.

8. Answer Key Item 8 [1 Mark]

Answer: 15 V

Teacher's Note:
a) Represents an induced potential difference or electromotive force.
b) Voltage magnitude is expressed in volts.

9. Answer Key Item 9 [2 Marks]

Answer: 1.31 m/s, 0.57 A

Teacher's Note:
a) Combines motional velocity and induced current values.
b) Check correct application of Faraday's and Ohm's laws.

10. Answer Key Item 10 [3 Marks]

Answer: \(\frac{7}{22}\) A, directed from E to A, \(\frac{3}{11}\) A, directed from B to E, \(\frac{1}{22}\) A, directed from F to E.

Teacher's Note:
a) Circuit branch currents are determined using Kirchhoff's laws.
b) Direction vectors for each branch current are explicitly specified.
c) Fractions should be kept in simplified form.

11. Answer Key Item 11 [1 Mark]

Answer: \(1.39 \times 10^{-5}\) V

Teacher's Note:
a) Small induced voltage calculated for electromagnetic induction problems.
b) Expressed cleanly in scientific notation.

12. Answer Key Item 12 [4 Marks]

Answer: \(3.94 \times 10^{-4}\) H, \(1.97 \times 10^{2}\) V, \(1.97 \times 10^{-2}\) Wb/sec, \(9.86 \times 10^{-5}\) coulomb

Teacher's Note:
a) Answers cover inductance, induced emf, rate of change of flux, and induced charge.
b) Multiple interrelated electromagnetic quantities are evaluated sequentially.
c) Check unit conversions for magnetic flux and inductance.
d) Precision must match standard physical constant rounding.

13. Answer Key Item 13 [1 Mark]

Answer: 640 Joule

Teacher's Note:
a) Total magnetic or electrical energy stored in the system.
b) Stored energy formula \(E = \frac{1}{2}LI^2\) or equivalent is applied.

14. Answer Key Item 14 [1 Mark]

Answer: 30 \(\mu\)V

Teacher's Note:
a) Micro-volt level induced emf.
b) Direct substitution into the rate of flux change formula.

15. Answer Key Item 15 [4 Marks]

Answer: (a) \(10^{4}\) As\(^{-1}\), (b) 0, (c) 2 A, (d) 173 \(\mu\)C

Teacher's Note:
a) Four distinct sub-quantities related to transient growth in inductive circuits.
b) Rate of change of current and instantaneous currents are analyzed.
c) Charge accumulation is computed by integrating current over time.
d) Check correct unit prefixes for micro-coulombs.

16. Answer Key Item 16 [1 Mark]

Answer: \(\frac{\mu_{0}aI}{2\pi R} \ln \frac{a+b}{b-a}\)

Teacher's Note:
a) Analytical expression for mutual inductance or magnetic flux linkage.
b) Integration over spatial dimensions yields the logarithmic term.

17. Answer Key Item 17 [1 Mark]

Answer: 1.25 A

Teacher's Note:
a) Steady-state or transient current magnitude.
b) Expressed in amperes.

18. Answer Key Item 18 [1 Mark]

Answer: \(6.25 \times 10^{-4}\) H

Teacher's Note:
a) Self-inductance value calculated in henries.
b) Proper power of ten formatting is essential.

19. Answer Key Item 19 [3 Marks]

Answer: (a) \(\Delta\phi = -36\mu\) weber; (b) \(\varepsilon = -\frac{d\phi}{dt} = 180\,\mu\)V; (c) yes

Teacher's Note:
a) Change in magnetic flux includes proper sign convention.
b) Induced emf calculation follows Faraday's law with negative sign.
c) Qualitative confirmation is given in part (c).

20. Answer Key Item 20 [1 Mark]

Answer: 0.252 \(\mu\)F

Teacher's Note:
a) Equivalent capacitance calculated for the network.
b) Units are correctly specified in micro-farads.