Physics: Electricity & Magnetism MCQs - 2

Question 1

A parallel plate capacitor with air between plates has capacitance C. If a dielectric slab of dielectric constant K = 4 and thickness t = d/2 (where d is plate separation) is inserted, the new capacitance becomes:

A) $\frac{4}{5}C$ B) $\frac{8}{5}C$ C) $2C$ D) $4C$

Correct Answer: B ($\frac{8}{5}C$)
Formula: $C’ = \frac{\epsilon_0 A}{d – t + \frac{t}{K}}$. Substituting $t = \frac{d}{2}$ and $K = 4$:
$C’ = \frac{\epsilon_0 A}{d – \frac{d}{2} + \frac{d}{8}} = \frac{\epsilon_0 A}{\frac{5d}{8}} = \frac{8}{5} \frac{\epsilon_0 A}{d} = \frac{8}{5}C$.

Question 2

A magnetic dipole of magnetic moment M is rotated through $180^\circ$ from the direction of a uniform magnetic field B. The work done in rotating the dipole is:

A) $-MB$ B) Zero C) $MB$ D) $2MB$

Correct Answer: D ($2MB$)
Explanation: Work done $W = -MB(\cos\theta_2 – \cos\theta_1)$. Here $\theta_1 = 0^\circ$ and $\theta_2 = 180^\circ$.
$W = -MB(\cos 180^\circ – \cos 0^\circ) = -MB(-1 – 1) = 2MB$.

Question 3

The self-inductance $L$ of a long solenoid having $N$ turns, length $l$, and cross-sectional area $A$ is directly proportional to:

A) $N$ B) $N^2$ C) $\frac{1}{N}$ D) $\sqrt{N}$

Correct Answer: B ($N^2$)
Explanation: Self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$. Hence, $L \propto N^2$.

Question 4

The Quality Factor ($Q$-factor) of a series LCR resonant circuit is expressed as:

A) $\frac{1}{R}\sqrt{\frac{L}{C}}$ B) $\frac{1}{L}\sqrt{\frac{R}{C}}$ C) $R\sqrt{\frac{C}{L}}$ D) $\frac{1}{C}\sqrt{\frac{L}{R}}$

Correct Answer: A ($\frac{1}{R}\sqrt{\frac{L}{C}}$)
Explanation: $Q = \frac{\omega_0 L}{R}$. Since resonant angular frequency $\omega_0 = \frac{1}{\sqrt{LC}}$, substituting gives $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$.

Question 5

A potentiometer wire of length 10 m has a resistance of 20 Ω. It is connected in series with a 3V battery and a resistance of 10 Ω. The potential gradient along the wire is:

A) 0.1 V/m B) 0.2 V/m C) 0.3 V/m D) 0.5 V/m

Correct Answer: B (0.2 V/m)
Calculation: Total resistance $R_{total} = 20 + 10 = 30\,\Omega$. Circuit current $I = \frac{3\text{ V}}{30\,\Omega} = 0.1\text{ A}$.
Voltage across wire $V_{wire} = I \times R_{wire} = 0.1 \times 20 = 2\text{ V}$.
Potential gradient $k = \frac{V_{wire}}{L} = \frac{2\text{ V}}{10\text{ m}} = 0.2\text{ V/m}$.

Question 6

What is the electric potential at any point on the equatorial plane of an electric dipole?

A) Maximum B) Dependent on dipole distance C) Zero D) Infinite

Correct Answer: C (Zero)
Explanation: Every point on the equatorial line is equidistant from both $+q$ and $-q$ charges of the dipole. The potential contributions $V_+ = \frac{kq}{r}$ and $V_- = -\frac{kq}{r}$ sum up to zero.

Question 7

A conducting rod of length $L$ rotates with constant angular velocity $\omega$ about one end in a uniform magnetic field $B$ perpendicular to the plane of rotation. The induced EMF between its ends is:

A) $BL\omega$ B) $\frac{1}{2}BL^2\omega$ C) $B^2 L \omega$ D) $\frac{1}{2}BL\omega^2$

Correct Answer: B ($\frac{1}{2}BL^2\omega$)
Explanation: Linear velocity varies from $0$ at the axis to $\omega L$ at the tip. Average linear speed $v_{avg} = \frac{\omega L}{2}$. Motional EMF $e = B \cdot v_{avg} \cdot L = \frac{1}{2} B L^2 \omega$.

Question 8

If the electric flux entering and leaving a closed surface are $\Phi_1$ and $\Phi_2$ respectively, the net charge enclosed inside the surface is:

A) $(\Phi_2 – \Phi_1)\epsilon_0$ B) $(\Phi_1 + \Phi_2)\epsilon_0$ C) $\frac{\Phi_2 – \Phi_1}{\epsilon_0}$ D) $\frac{\epsilon_0}{\Phi_2 – \Phi_1}$

Correct Answer: A ($(\Phi_2 – \Phi_1)\epsilon_0$)
Explanation: Net outward electric flux $\Phi_{net} = \Phi_2 – \Phi_1$. By Gauss’s Law, $\Phi_{net} = \frac{q_{enclosed}}{\epsilon_0} \implies q_{enclosed} = (\Phi_2 – \Phi_1)\epsilon_0$.

Question 9

A step-up transformer has a turn ratio of $1:10$. If an alternating voltage of $220\text{ V}$ is applied across the primary winding, the secondary voltage will be:

A) 22 V B) 110 V C) 220 V D) 2200 V

Correct Answer: D (2200 V)
Explanation: Transformer ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p} = 10 \implies V_s = 10 \times 220\text{ V} = 2200\text{ V}$.

Question 10

The frequency of gyration (cyclotron frequency) of a charged particle moving in a uniform magnetic field is independent of its:

A) Mass B) Charge C) Speed D) Magnetic field strength

Correct Answer: C (Speed)
Explanation: Cyclotron frequency $f = \frac{qB}{2\pi m}$. It depends on $q$, $B$, and $m$, but is independent of the particle’s speed or radius of orbit.

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Physics: Electricity & Magnetism MCQs – 2

Physics: Electricity & Magnetism (Part 2)

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Physics: Electricity & Magnetism (Part 2)

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