Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
When a coil is placed in a time dependent magnetic field the power dissipated in it is $P$. The number of turns, area of the coil and radius of the coil wire are $N$, $A$ and $r$ respectively. For a second coils number of turns, area of the coil and radius of the coil wire are $2N$, $2A$ and $3r$ respectively. When the first coil is replaced with second coil the power dissipated in it is $\sqrt{2}\,\alpha P$. The value of $\alpha$ is ______.
(1) $36$
(2) $128\sqrt{2}$
(3) $16$
(4) $64$
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Answer: Option (1)
Key idea: Electromagnetic Induction: Power dissipation in coils with varying parameters (turns, area, wire radius)
Induced EMF in a coil: $\varepsilon = NA\frac{dB}{dt}$. For coil 2: $\varepsilon_2 = 2N \cdot 2A \cdot \frac{dB}{dt} = 4\varepsilon_1$
Resistance of coil wire: $R = \rho \frac{L}{A_{wire}}$ where $L = N \times \text{perimeter} \propto N\sqrt{A}$ and $A_{wire} = \pi r^2$
For coil 1: $R_1 \propto \frac{N\sqrt{A}}{r^2}$. For coil 2: $R_2 \propto \frac{2N\sqrt{2A}}{(3r)^2} = \frac{2\sqrt{2}N\sqrt{A}}{9r^2}$
Ratio: $\frac{R_2}{R_1} = \frac{2\sqrt{2}}{9}$
Power dissipated: $P = \frac{\varepsilon^2}{R}$, so $\frac{P_2}{P_1} = \left(\frac{\varepsilon_2}{\varepsilon_1}\right)^2 \cdot \frac{R_1}{R_2} = 16 \times \frac{9}{2\sqrt{2}} = \frac{72}{\sqrt{2}} = 36\sqrt{2}$
Given $P_2 = \sqrt{2}\alpha P_1$, we get $\sqrt{2}\alpha = 36\sqrt{2}$, therefore $\alpha = 36$
A circular current loop of radius $R$ is placed inside square loop of side length $L$ $(L >> R)$ such that they are co-planar and their centers coincide. The permeability of free space is $\mu_0$. The mutual inductance between circular loop and square loop is ________.
(1) $2\sqrt{2}\frac{\mu_0 L^2}{R}$
(2) $\sqrt{2}\frac{\mu_0 L^2}{R}$
(3) $\sqrt{2}\frac{\mu_0 R^2}{L}$
(4) $2\sqrt{2}\frac{\mu_0 R^2}{L}$
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Answer: Option (4)
Key idea: Mutual Inductance and Magnetic Field of Square Loop
Find the magnetic field at the center of a square loop of side $L$ carrying current $I$. Each side at distance $d = L/2$ contributes: $B_{side} = \frac{\mu_0 I}{4\pi d} \times 2\sin(45°) = \frac{\sqrt{2}\mu_0 I}{2\pi L}$
Total field from all 4 sides (perpendicular to plane): $B_{center} = 4 \times \frac{\sqrt{2}\mu_0 I}{2\pi L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}$
Since $L >> R$, the field is approximately uniform over the circular loop area. Calculate flux: $\Phi = B \times \pi R^2 = \frac{2\sqrt{2}\mu_0 I}{\pi L} \times \pi R^2$
Simplify the flux: $\Phi = \frac{2\sqrt{2}\mu_0 I R^2}{L}$
In the given circuit below inductance values of $L_1, L_2$ and $L_3$ are same. The magnetic energy stored in the entire circuit is $(U_L)$ and that stored in the $L_2$ inductor is $(U_l)$. $U_L / U_l$ is ______.
(Ignore the mutual inductance if any)
[Figure: Circuit diagram showing three inductors: $L_1$ in series with a parallel combination of $L_2$ and $L_3$]
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Answer: 6
Key idea: Energy stored in inductors in series-parallel circuits
Given: $L_1 = L_2 = L_3 = L$, with $L_1$ in series with parallel combination of $L_2$ and $L_3$
Let total current be $I$. Current through $L_1$ is $I$, and since $L_2 = L_3$, current splits equally: $I_2 = I_3 = I/2$
Energy in $L_2$: $U_l = \frac{1}{2}L(I/2)^2 = \frac{LI^2}{8}$
Energy in $L_1$: $U_1 = \frac{1}{2}LI^2 = \frac{LI^2}{2}$; Energy in $L_3$: $U_3 = \frac{LI^2}{8}$
A metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r) = B_o e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:
An inductor of inductance 10 mH having resistance of 100 $\Omega$ is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 2 mA and 4 mA is _______. [Figure: Circuit diagram showing an inductor in series with a capacitor and a switch, connected to a battery]
(1) $4/3$
(2) $3/4$
(3) $5/3$
(4) $3/5$
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Answer: Option (1)
Key idea: LR Circuit - Voltage distribution using Kirchhoff's Voltage Law
Given: $L = 10$ mH, $R = 100\,\Omega$, $E = 1.0$ V, find ratio of $V_L$ at $i_1 = 2$ mA and $i_2 = 4$ mA
Apply Kirchhoff's Voltage Law: $E = iR + V_L$, so $V_L = E - iR$
When $i_1 = 2$ mA: $V_{L1} = 1.0 - (0.002)(100) = 1.0 - 0.2 = 0.8$ V
When $i_2 = 4$ mA: $V_{L2} = 1.0 - (0.004)(100) = 1.0 - 0.4 = 0.6$ V
A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as $B=0.4 \sin(300t)$ Tesla. The normal to the plane of loop makes an angle of $60^\circ$ with the field. The maximum induced emf produced in the loop is __________ mV.
(1) 12
(2) 18
(3) 21
(4) 24
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Answer: Option (4)
Key idea: Faraday's Law of Electromagnetic Induction
Calculate area of square loop: $A = (0.02)^2 = 4 \times 10^{-4}$ m²
A 30 cm long solenoid has 10 turns per cm and area of 5 cm$^2$. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is $\alpha \times 10^{-5}$ V. The value of $\alpha$ is _______.
(1) 60
(2) 12
(3) 120
(4) 34
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Answer: Option (2)
Key idea: Self-Inductance and Faraday's Law of Electromagnetic Induction
