Physics 2 Quiz: Multi Step Eandm Problems

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Multi Step Eandm ProblemsQuestion 1 of 10

A point charge $+Q$ is placed at the center of a thick spherical conducting shell with inner radius $a$ and outer radius $b$ . The shell carries a net charge of $-3Q$ .

What is the surface charge density on the outer surface of the shell, and what is the direction of the electric field in the region $r > b$ ?

Outer surface charge density

\sigma_{outer} = \frac{-2Q}{4\pi b^2}

; field points radially inward (toward the center) for all

r > b

, because the net charge enclosed is negative.

Outer surface charge density

\sigma_{outer} = \frac{-3Q}{4\pi b^2}

; field points radially inward for all

r > b

, because the shell's charge dominates and the enclosed net charge is

-3Q

Outer surface charge density

\sigma_{outer} = \frac{-2Q}{4\pi b^2}

; field points radially outward for all

r > b

, because the central charge dominates over the shell charge in the exterior region.

Outer surface charge density

\sigma_{outer} = \frac{+2Q}{4\pi b^2}

; field points radially outward for all

r > b

, because the inner surface holds

-Q

leaving

-2Q

on the outer surface, which then combines with the central

+Q

to give a net positive exterior charge.

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Physics 2 Quiz

Physics 2 Quiz: Multi Step Eandm Problems

Practice Multi Step Eandm Problems in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Multi Step Eandm Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A point charge $+Q$ is placed at the center of a thick spherical conducting shell with inner radius $a$ and outer radius $b$ . The shell carries a net charge of $-3Q$ .

What is the surface charge density on the outer surface of the shell, and what is the direction of the electric field in the region $r > b$ ?

Outer surface charge density $\sigma_{outer} = \frac{-2Q}{4\pi b^2}$ ; field points radially inward (toward the center) for all $r > b$ , because the net charge enclosed is negative. (correct answer)
Outer surface charge density $\sigma_{outer} = \frac{-3Q}{4\pi b^2}$ ; field points radially inward for all $r > b$ , because the shell's charge dominates and the enclosed net charge is $-3Q$ .
Outer surface charge density $\sigma_{outer} = \frac{-2Q}{4\pi b^2}$ ; field points radially outward for all $r > b$ , because the central charge dominates over the shell charge in the exterior region.
Outer surface charge density $\sigma_{outer} = \frac{+2Q}{4\pi b^2}$ ; field points radially outward for all $r > b$ , because the inner surface holds $-Q$ leaving $-2Q$ on the outer surface, which then combines with the central $+Q$ to give a net positive exterior charge.

Explanation: Whenever you see a conducting shell with charges, your first move should be Gauss's Law combined with the rule that the electric field inside a conductor is zero. That zero-field condition is what forces charge to redistribute on the inner and outer surfaces. Here's the logic chain: The point charge

+Q

at the center induces

-Q

on the inner surface (at radius

a

) to cancel the field inside the conducting material. Since the shell carries a net charge of

-3Q

total, and

-Q

is already "used up" on the inner surface, the remaining

-3Q - (-Q) = -2Q

must reside on the outer surface. This gives an outer surface charge density of

\sigma_{outer} = \frac{-2Q}{4\pi b^2}

. Now apply Gauss's Law for

r > b

: the total enclosed charge is

+Q + (-Q) + (-2Q) = -2Q

, which is negative, so the electric field points radially inward. That's answer A. Choice B is tempting but wrong — it ignores the inner surface induction entirely and treats the full

-3Q

as sitting on the outer surface, violating charge conservation on the shell. Choice C correctly identifies the outer surface charge density but gets the field direction wrong; a net enclosed charge of

-2Q

absolutely produces an inward-pointing field — the "central charge dominates" reasoning is fabricated and has no physical basis. Choice D reverses the sign of the outer charge; the outer surface holds

-2Q

, not

+2Q

. Study tip: Always track charge in two steps — inner surface induction first (

-Q_{center}

), then subtract from the shell's net charge to find the outer surface charge. This two-step method prevents all the errors seen in B, C, and D.

Question 2

An RLC series circuit consists of a resistor $R = 100\,\Omega$ , an inductor $L = 0.1\,\text{H}$ , and a capacitor $C = 10\,\mu\text{F}$ , driven by an AC source $V(t) = V_0 \cos(\omega t)$ . The source frequency is set to twice the resonant angular frequency: $\omega = 2\omega_0$ , where $\omega_0 = 1/\sqrt{LC}$ .

At this driving frequency $\omega = 2\omega_0$ , which of the following correctly characterizes the phase relationship between the current and the voltage source, and the impedance of the circuit?

The current lags the voltage because the inductive reactance exceeds the capacitive reactance at $\omega > \omega_0$ ; the impedance is $Z = \sqrt{R^2 + \left(\frac{3\omega_0 L}{2}\right)^2}$ , since $X_L - X_C = 2\omega_0 L - \dfrac{\omega_0 L}{2} = \dfrac{3\omega_0 L}{2}$ . (correct answer)
The current leads the voltage because doubling the frequency means the capacitor dominates; the impedance is $Z = \sqrt{R^2 + \left(\frac{3\omega_0 L}{2}\right)^2}$ , since $X_C > X_L$ above the resonant frequency.
The current lags the voltage because the inductive reactance exceeds the capacitive reactance; the impedance is $Z = \sqrt{R^2 + (2\omega_0 L)^2}$ , because the capacitive reactance becomes negligible at high frequencies and can be omitted.
The current is in phase with the voltage because the circuit is being driven at a harmonic of the resonant frequency, and harmonics also produce resonance conditions; the impedance equals $R$ at all integer multiples of $\omega_0$ .

Explanation: When analyzing an RLC series circuit driven off-resonance, your goal is to compare inductive and capacitive reactances at the given frequency, then calculate impedance from their net difference. At resonance,

\omega_0 = 1/\sqrt{LC}

, which gives us a useful identity:

\omega_0^2 LC = 1

, so

\omega_0 L = 1/(\omega_0 C)

. When the driving frequency is

\omega = 2\omega_0

, the reactances become

X_L = \omega L = 2\omega_0 L

and

X_C = 1/(\omega C) = 1/(2\omega_0 C) = \omega_0 L/2

. Since

X_L > X_C

, the circuit is net inductive — meaning the current lags the voltage. The net reactance is

X_L - X_C = 2\omega_0 L - \frac{\omega_0 L}{2} = \frac{3\omega_0 L}{2}

, giving impedance

Z = \sqrt{R^2 + \left(\frac{3\omega_0 L}{2}\right)^2}

. This confirms answer A is correct. Answer B makes a critical error: it claims the capacitor dominates above resonance, but the opposite is true — inductive reactance grows with frequency while capacitive reactance shrinks. The impedance formula in B happens to be correct, but the phase reasoning is backwards. Answer C correctly identifies the lag but drops the capacitive term entirely, which is only valid at extremely high frequencies. At

\omega = 2\omega_0

X_C

is still significant and cannot be ignored — omitting it gives the wrong impedance. Answer D confuses harmonics with resonance. True resonance only occurs at

\omega_0

; driving at integer multiples does not reproduce the resonance condition. Study tip: Always calculate both reactances explicitly before concluding which dominates — never assume based on frequency alone.

Question 3

A coaxial cable consists of a solid inner conductor of radius $a$ carrying current $I$ uniformly distributed over its cross-section, surrounded by a thin outer cylindrical shell of radius $b$ carrying current $I$ in the opposite direction. The space between the conductors is vacuum.

At a radial distance $r$ such that $a < r < b$ , what is the magnitude of the magnetic field, and how does the energy stored per unit length in the magnetic field in this region scale with $I$ ?

$B = 0$ for $a < r < b$ because the inner and outer currents are equal and opposite, and by Gauss's law for magnetism, the fields cancel in the region between conductors, just as they do outside the cable.
$B = \frac{\mu_0 I r}{2\pi a^2}$ ; the energy per unit length scales as $I^2$ because the field between the conductors still depends linearly on $I$ , giving energy per unit length $\propto I^2(b^2 - a^2)/a^4$ .
$B = \frac{\mu_0 I}{2\pi r}$ ; the energy per unit length scales as $I$ because the energy density is proportional to $B$ , not $B^2$ , and $B \propto I$ , yielding a linear dependence on current.
$B = \frac{\mu_0 I}{2\pi r}$ ; the energy per unit length scales as $I^2$ because $u = B^2/(2\mu_0)$ and $B \propto I$ , so integrating over the annular region gives energy per unit length $\propto I^2 \ln(b/a)$ . (correct answer)

Explanation: Whenever you see a question combining Ampère's Law with energy storage, treat them as two separate sub-problems: first find

B

, then apply the correct energy density formula. For the region

a < r < b

, draw an Amperian loop of radius

r

. The only current enclosed is the full inner current

I

— the outer shell at radius

b

lies entirely outside your loop and contributes nothing. Ampère's Law gives:

B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}

Now for energy: the magnetic energy density is

u = \frac{B^2}{2\mu_0}

, so integrating over the annular cross-section yields energy per unit length:

\frac{U}{\ell} = \int_a^b \frac{B^2}{2\mu_0}(2\pi r)\,dr = \frac{\mu_0 I^2}{4\pi}\ln\!\left(\frac{b}{a}\right)

This scales as

I^2

, confirming D is correct. A is wrong on two counts: it misapplies "Gauss's law for magnetism" (which says no magnetic monopoles, not that fields cancel) and incorrectly treats the geometry like the exterior of a coaxial cable, where both currents are enclosed. Outside

r > b

, the fields cancel — not between the conductors. B uses the field formula for inside the solid conductor (

r < a

), where current is distributed over the cross-section. Between the conductors, all of

I

is enclosed, so

B = \mu_0 I / 2\pi r

, not

\mu_0 I r / 2\pi a^2

. C gets

B

right but claims energy scales as

I

rather than

I^2

. Energy density goes as

B^2

, never

B

alone — this is a classic trap. Study tip: Always ask "what current does my Amperian loop enclose?" separately from "what is the energy density formula?" — these are two independent steps that must both be correct.

Question 4

A uniform electric field $\mathbf{E} = E_0\,\hat{z}$ exists in a region of space. A dipole consisting of charges $+q$ and $-q$ separated by distance $d$ is placed in this field with the dipole moment $\mathbf{p} = qd\,\hat{x}$ (perpendicular to $\mathbf{E}$ ). The dipole is then released from rest.

Immediately after release, what is the net force on the dipole and what is the net torque about the dipole's center?

Net force: $qE_0\,\hat{z}$ (upward on the dipole); net torque: zero, because the uniform field exerts equal forces on $+q$ and $-q$ that together produce a net translational push while their torque contributions cancel by symmetry.
Net force: zero; net torque: $\tau = pE_0$ , directed so as to rotate $\mathbf{p}$ toward alignment with $\mathbf{E}$ , because a uniform field exerts no net translational force on a dipole but does exert a torque $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$ . (correct answer)
Net force: zero; net torque: zero, because $\mathbf{p}$ is perpendicular to $\mathbf{E}$ and the torque magnitude $|\mathbf{p}\times\mathbf{E}| = pE_0\sin\theta$ vanishes when $\theta = 90°$ .
Net force: $2qE_0\,\hat{z}$ (upward); net torque: $pE_0$ directed to rotate $\mathbf{p}$ toward $\mathbf{E}$ , because each charge experiences force $qE_0\,\hat{z}$ and the torques from the two charges add constructively when $\mathbf{p} \perp \mathbf{E}$ .

Explanation: When you encounter a dipole in an electric field, train yourself to ask two separate questions: Is the field uniform? and What is the angle between p and E? These two facts determine everything. In a uniform electric field, the force on

+q

+qE_0\hat{z}

and on

-q

-qE_0\hat{z}

. These are equal and opposite, so they cancel perfectly — the net force is always zero for a dipole in a uniform field, regardless of orientation. For torque, you use

\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}

. Here

\mathbf{p} = qd\,\hat{x}

and

\mathbf{E} = E_0\hat{z}

, so

|\boldsymbol{\tau}| = pE_0\sin\theta

. With

\theta = 90°

between

\hat{x}

and

\hat{z}

, you get

|\boldsymbol{\tau}| = pE_0

— maximum torque, directed to rotate p toward alignment with E. That makes B correct. A is wrong because it invents a net translational force. The equal-and-opposite forces on the two charges cancel — they cannot "add up" to push the dipole anywhere. C contains a critical math error:

\sin(90°) = 1

, not zero. The torque is actually maximum when p ⊥ E, not zero. Students sometimes confuse this with potential energy (

U = -\mathbf{p}\cdot\mathbf{E}

), which is zero at 90°. D double-counts the forces (the two

qE_0

forces oppose each other, not add) while accidentally getting the torque magnitude right for the wrong reasons. Your study tip: memorize the two key results — uniform field means zero net force always, and torque is maximum (not zero) when the dipole is perpendicular to the field.

Question 5

A long solenoid of radius $R$ , $n$ turns per unit length, carries a current $I(t) = I_0 \sin(\omega t)$ . A circular conducting loop of radius $r < R$ lies in the plane perpendicular to the solenoid axis and is coaxial with it. The loop has resistance $\mathcal{R}$ and negligible self-inductance.

What is the amplitude of the current induced in the small loop, and in what direction does it flow relative to the solenoid current at the moment when $I(t)$ is at its maximum positive value?

Amplitude $\frac{\mu_0 n \omega I_0 \pi R^2}{\mathcal{R}}$ ; at the moment $I$ is maximum, the induced current is also at its maximum amplitude and flows in the same direction as the solenoid current to reinforce the flux.
Amplitude $\frac{\mu_0 n \omega I_0 \pi r^2}{\mathcal{R}}$ ; at the moment $I$ is maximum, the induced EMF is also maximum and the induced current flows opposite to the solenoid current to oppose the large flux.
Amplitude $\frac{\mu_0 n \omega I_0 \pi R^2}{\mathcal{R}}$ ; at the moment $I$ is maximum, the induced current is zero because the EMF is proportional to $\cos(\omega t)$ , which equals zero when $\sin(\omega t) = 1$ .
Amplitude $\frac{\mu_0 n \omega I_0 \pi r^2}{\mathcal{R}}$ ; at the moment $I$ is maximum, the induced current is zero because the EMF is proportional to $\cos(\omega t)$ , which equals zero when $\sin(\omega t) = 1$ . (correct answer)

Explanation: Whenever you see a question about electromagnetic induction with a time-varying current, your first instinct should be Faraday's Law: the induced EMF depends on the rate of change of flux, not the flux itself. Here's the full reasoning. The magnetic field inside the solenoid is

B(t) = \mu_0 n I_0 \sin(\omega t)

. Since the small loop (radius

r < R

) lies entirely inside the solenoid, the flux through it uses the loop's own area:

\Phi = \mu_0 n I_0 \sin(\omega t) \cdot \pi r^2

. Applying Faraday's Law, the induced EMF is

\mathcal{E} = -\frac{d\Phi}{dt} = -\mu_0 n I_0 \omega \pi r^2 \cos(\omega t)

. The amplitude is therefore

\frac{\mu_0 n \omega I_0 \pi r^2}{\mathcal{R}}

after dividing by resistance. Crucially, at the moment

I

is maximum,

\sin(\omega t) = 1

, which means

\cos(\omega t) = 0

. The EMF — and thus the induced current — is zero at that instant. This confirms D. Choice A is wrong on both counts: it uses

R

(solenoid radius) instead of

r

, and incorrectly claims the induced current is maximum and co-directional when

I

peaks. Choice B uses the correct area

\pi r^2

but makes the same timing error — claiming EMF is maximum when

I

is maximum, which confuses flux with its derivative. Choice C gets the timing right (induced current is zero when

\sin(\omega t) = 1

) but uses the wrong area

\pi R^2

instead of

\pi r^2

. Study tip: Always distinguish between the area that captures flux (the loop's own area when fully inside the solenoid) and the solenoid's cross-section. And remember: EMF peaks when current crosses zero, not when it peaks.

Question 6

Two identical resistors, each of resistance $R$ , are connected in series with an ideal inductor of inductance $L$ and an ideal battery of EMF $\mathcal{E}$ . The circuit has been running for a very long time. One of the resistors is then suddenly short-circuited (replaced by a wire) at time $t = 0$ .

Immediately after the short circuit is applied ( $t = 0^+$ ), what is the current through the inductor, and what is the voltage across the remaining (non-shorted) resistor?

Current: $\mathcal{E}/(2R)$ ; voltage: $\mathcal{E}/2$ . The inductor maintains the pre-short-circuit current, and with only one resistor now in the loop, the full inductor current flows through it, giving voltage $I_0 R = \mathcal{E}/2$ . (correct answer)
Current: $\mathcal{E}/R$ ; voltage: $\mathcal{E}$ . Immediately after the short, the circuit reconfigures instantly so the current jumps to the new DC steady state, doubling because the total resistance is halved.
Current: $\mathcal{E}/(2R)$ ; voltage: $\mathcal{E}$ . The inductor maintains its pre-short-circuit current, but the voltage across the remaining resistor equals the battery EMF because Kirchhoff's voltage law requires the full EMF to appear across the single remaining resistor.
Current: $\mathcal{E}/(2R)$ ; voltage: $0$ . The inductor maintains its pre-short-circuit current, but the voltage across any resistor in a circuit with an inductor is zero immediately after a switching event because the inductor absorbs all the voltage transiently.

Explanation: When a circuit contains an inductor, your first instinct should be to apply the inductor's fundamental constraint: current through an inductor cannot change instantaneously. This single rule unlocks the entire problem. Before the short circuit, both resistors are in series with the battery, so the steady-state current is

I_0 = \mathcal{E}/(2R)

. At

t = 0^+

, the inductor "remembers" this current — it immediately freezes at

\mathcal{E}/(2R)

regardless of what the rest of the circuit is doing. That confirms the first part of answer A. Now apply Kirchhoff's Voltage Law around the new loop (battery, inductor, and the one surviving resistor — the shorted resistor contributes zero voltage). With current

\mathcal{E}/(2R)

flowing through resistance

R

, the voltage across the remaining resistor is simply:

V_R = I_0 \cdot R = \frac{\mathcal{E}}{2R} \cdot R = \frac{\mathcal{E}}{2}

This confirms A as correct. B is wrong because it violates the inductor's core constraint — current cannot jump instantaneously to a new steady state. C is tempting but contains a subtle error: if the current is

\mathcal{E}/(2R)

and the resistor is

R

, Ohm's Law gives

\mathcal{E}/2

, not

\mathcal{E}

. Claiming the full EMF appears across the resistor would require a different (larger) current. D is wrong because the inductor constraining current doesn't mean voltage across resistors drops to zero — Ohm's Law still applies to resistors at every instant. Study tip: In inductor switching problems, always find the pre-switch current first, freeze it at

t = 0^+

, then redraw the circuit and apply KVL/Ohm's Law normally. Never let the new topology change your current instantaneously.

Question 7

An ideal transformer has a primary coil with $N_p$ turns connected to an AC source of RMS voltage $V_p$ , and a secondary coil with $N_s = 3N_p$ turns connected to a purely resistive load $R_L$ . The transformer is ideal (100% efficiency). A second identical resistor $R_L$ is then connected in parallel with the first across the secondary terminals.

After the second resistor is connected in parallel, compared to the original single-load configuration, how do the RMS secondary voltage and the RMS primary current change?

The secondary voltage remains $3V_p$ and the primary current remains unchanged, because an ideal transformer adjusts the secondary current to maintain constant voltage, so the primary side is unaffected by changes in load.
The secondary voltage decreases to $3V_p/2$ because the parallel resistors draw more current and the transformer cannot maintain the full step-up ratio under higher load; the primary current increases by a factor of $\sqrt{2}$ .
The secondary voltage remains $3V_p$ because it is set by the turns ratio regardless of load; the primary current doubles because power delivered to the secondary doubles when the load resistance is halved, and $P = V_p I_p$ . (correct answer)
The secondary voltage remains $3V_p$ because the turns ratio is fixed; the primary current increases by a factor of $\sqrt{2}$ because the secondary current increases by $\sqrt{2}$ when two equal resistors share the load, and the transformer reflects this increase to the primary.

Explanation: Whenever you see a transformer question, anchor yourself to two governing equations: the voltage ratio

\frac{V_s}{V_p} = \frac{N_s}{N_p}

and the power conservation condition

P_p = P_s

, which gives

V_p I_p = V_s I_s

. For an ideal transformer, the secondary voltage is determined entirely by the turns ratio and the primary voltage — it does not depend on what load is connected. With

N_s = 3N_p

, the secondary voltage is fixed at

V_s = 3V_p

regardless of whether one or two resistors are connected. When you add a second

R_L

in parallel, the equivalent load resistance drops from

R_L

R_L/2

. Since

V_s

is unchanged, the secondary current doubles:

I_s = V_s / R_{eq}

goes from

3V_p/R_L

6V_p/R_L

. The power delivered to the secondary therefore doubles. By conservation of energy, the primary must supply twice the power, and since

V_p

is fixed by the AC source,

I_p

must double. This confirms C is correct. A is wrong because it concludes the primary current is unchanged — that would violate energy conservation, since more power is being delivered to the load. B is wrong on two counts: secondary voltage in an ideal transformer doesn't drop with increased load (that only happens in real transformers with winding resistance), and the

\sqrt{2}

factor has no physical basis here. D is tempting but contains a key error: secondary current doubles (not by

\sqrt{2}

) because the resistance is halved, not shared equally in some root-mean sense. Study tip: On transformer problems, always ask two questions in order — (1) what sets the voltage? (turns ratio), then (2) what sets the current? (the load and power conservation). Never let the load affect the voltage in an ideal transformer.

Question 8

A long straight wire carries a steady current $I_1$ in the $+\hat{z}$ direction. A rectangular conducting loop of width $w$ and height $h$ lies in the $xz$ -plane with its near edge at distance $d$ from the wire and its far edge at distance $d + w$ . The loop has resistance $\mathcal{R}$ and is pulled away from the wire (in the $+\hat{x}$ direction) with constant velocity $v$ .

As the loop is pulled away from the wire at constant velocity, which of the following correctly describes the induced EMF and the direction of the induced current in the loop?

EMF: $\mathcal{E} = \frac{\mu_0 I_1 h v}{2\pi} \cdot \frac{w}{d(d+w)}$ ; current flows clockwise (when viewed from $+\hat{y}$ ) to oppose the motion of the loop, consistent with Lenz's law.
EMF: $\mathcal{E} = \frac{\mu_0 I_1 h v}{2\pi} \cdot \frac{1}{d}$ ; current flows clockwise (when viewed from $+\hat{y}$ ) to reinforce the decreasing flux through the loop, consistent with Lenz's law.
EMF: $\mathcal{E} = \frac{\mu_0 I_1 h v}{2\pi} \cdot \frac{w}{d(d+w)}$ ; current flows counterclockwise (when viewed from $+\hat{y}$ ) to oppose the decreasing flux through the loop. (correct answer)
EMF: $\mathcal{E} = \frac{\mu_0 I_1 h v}{2\pi} \cdot \frac{1}{d}$ ; current flows counterclockwise (when viewed from $+\hat{y}$ ) to oppose the decreasing flux through the loop, consistent with Lenz's law.

Explanation: Whenever you see a loop moving near a current-carrying wire, your two jobs are: (1) calculate the EMF using Faraday's law, and (2) determine current direction using Lenz's law. Calculating the EMF. The magnetic field from the wire at distance

x

B(x) = \frac{\mu_0 I_1}{2\pi x}

, directed in

-\hat{y}

inside the loop (by the right-hand rule, since current flows in

+\hat{z}

). The flux through the loop is

\Phi = \int_d^{d+w} \frac{\mu_0 I_1 h}{2\pi x}\,dx = \frac{\mu_0 I_1 h}{2\pi}\ln\!\left(\frac{d+w}{d}\right)

. Taking the time derivative and using

\dot{d} = v

, the induced EMF becomes

\mathcal{E} = \frac{\mu_0 I_1 h v}{2\pi}\cdot\frac{w}{d(d+w)}

, which matches answer C. Applying Lenz's law. As the loop moves away,

\Phi

(in

-\hat{y}

) decreases. To oppose this decrease, the induced current must create flux in

-\hat{y}

inside the loop — meaning the current flows counterclockwise when viewed from

+\hat{y}

. This confirms C. Why the others fail. Answer A gets the EMF formula right but incorrectly states the current is clockwise — clockwise would reinforce the decrease, violating Lenz's law. Answers B and D both use the wrong EMF formula

\frac{1}{d}

, which would only be valid if the loop were infinitesimally thin; integrating across the loop's width

w

is essential and produces the

\frac{w}{d(d+w)}

factor. D also gets the current direction wrong. Study tip: Always integrate

B

across the loop's width — never just evaluate it at the near edge. And remember: Lenz's law opposes change in flux, not motion directly.

Question 9

A parallel-plate capacitor with plate area $A$ and separation $d$ is fully charged by a battery of EMF $\mathcal{E}$ and then disconnected. A dielectric slab with dielectric constant $\kappa$ is then inserted to fill the gap completely.

After the dielectric is inserted, which of the following correctly describes the changes in both the electric field between the plates and the energy stored in the capacitor?

The electric field decreases by a factor of $\kappa$ because the bound surface charges on the dielectric partially cancel the free charges, and the stored energy decreases by a factor of $\kappa$ because $U = Q^2/(2C)$ and $C$ increases while $Q$ remains fixed. (correct answer)
The electric field remains unchanged because the charge $Q$ on the plates is fixed and Gauss's law requires the field to depend only on free charge, and the stored energy increases by a factor of $\kappa$ because the dielectric adds polarization energy to the system.
The electric field decreases by a factor of $\kappa$ because the bound surface charges partially cancel the free charges, and the stored energy increases by a factor of $\kappa$ because the capacitance increases and $U = Q^2/(2C)$ grows with $C$ .
The electric field decreases by a factor of $\kappa$ and the stored energy remains unchanged because the work done by the electric field on the dielectric exactly compensates for the reduction in field energy density.

Explanation: When a capacitor is disconnected from its battery before a dielectric is inserted, the key constraint is that the charge

Q

on the plates is fixed — no charge can flow on or off. This single fact determines everything else. Here's the reasoning: the dielectric becomes polarized, creating bound surface charges that oppose the free charges on the plates. By Gauss's law applied to the full field (including polarization), the net field inside decreases:

E = E_0/\kappa

, where

E_0 = \mathcal{E}/d

was the original field. Meanwhile, inserting the dielectric increases the capacitance to

C' = \kappa C_0

. Since

Q

is fixed, the stored energy becomes

U' = Q^2/(2C') = Q^2/(2\kappa C_0) = U_0/\kappa

. Both quantities — field and energy — decrease by the same factor

\kappa

. This is exactly what answer A describes, making it correct. Answer B is wrong on both counts. Gauss's law with only free charges gives

D

, not

E

— the bound charges absolutely affect

E

. Energy does not increase; it decreases. Answer C correctly identifies the field reduction but gets the energy direction backwards.

U = Q^2/(2C)

decreases as

C

increases — you're dividing by a larger number, not multiplying. Many students confuse this with the constant-voltage case (

U = \frac{1}{2}C V^2

), where energy would increase. Answer D is a fabricated-sounding statement with no physical basis; energy is not conserved within the capacitor alone here — it's transferred to mechanical work pulling the dielectric in. Study tip: Always identify whether the battery is connected (voltage fixed) or disconnected (charge fixed) before analyzing a dielectric insertion — the two scenarios give opposite energy trends.

Question 10

A proton moves with velocity $\mathbf{v} = v_0\,\hat{x}$ into a region where a uniform electric field $\mathbf{E} = E_0\,\hat{y}$ and a uniform magnetic field $\mathbf{B} = B_0\,\hat{z}$ exist simultaneously. The proton's initial kinetic energy is $K_0$ .

For the proton to travel through this region in a straight line at constant speed, what condition must hold, and which field (if either) does net work on the proton during this straight-line motion?

The condition $v_0 = E_0/B_0$ must hold so that the electric and magnetic forces cancel; the electric field does positive work on the proton, while the magnetic field does equal negative work, so the net work is zero and kinetic energy is conserved.
The condition $v_0 = E_0/B_0$ must hold so that the electric and magnetic forces cancel; neither field does work on the proton because the electric force is perpendicular to the proton's displacement and the magnetic force is always perpendicular to the velocity. (correct answer)
The condition $v_0 = E_0/B_0$ must hold so that the electric and magnetic forces cancel; the electric field does work $qE_0 \Delta x$ on the proton over displacement $\Delta x$ in the $\hat{x}$ -direction, while the magnetic force does no work, resulting in a net energy gain.
The condition $v_0 = B_0/E_0$ must hold to balance the forces; the magnetic field does positive work because the Lorentz force has a component along the velocity, and the electric field does equal negative work to maintain constant speed.

Explanation: When a charged particle moves through crossed electric and magnetic fields, you're dealing with a velocity selector — a classic setup where force balance determines straight-line motion. For a proton moving in

\hat{x}

with

\mathbf{E} = E_0\hat{y}

and

\mathbf{B} = B_0\hat{z}

, the electric force is

qE_0\hat{y}

and the magnetic force is

q\mathbf{v}\times\mathbf{B} = qv_0B_0(\hat{x}\times\hat{z}) = -qv_0B_0\hat{y}

. For these to cancel, you need

qE_0 = qv_0B_0

, giving the selector condition

v_0 = E_0/B_0

. So far, every answer agrees on this part — the real trap is the work question. B is correct because once you identify that the net force is zero and the proton travels in a straight line along

\hat{x}

, neither field does work. The electric force points in

\hat{y}

, which is perpendicular to the displacement

\Delta x\,\hat{x}

, so

W_E = \mathbf{F}_E \cdot \Delta\mathbf{x} = 0

. The magnetic force is always perpendicular to velocity by definition, so

W_B = 0

always. Net work is zero and kinetic energy is conserved — not because two nonzero works cancel, but because both are independently zero. A is tempting but wrong: it claims the electric field does positive work while the magnetic field does negative work. Magnetic forces never do work — this is a fundamental fact, not a coincidence of cancellation. C incorrectly computes

W_E = qE_0\Delta x

, forgetting that the electric force is in

\hat{y}

while displacement is in

\hat{x}

— they're perpendicular, so the dot product is zero. D inverts the selector condition to

B_0/E_0

and falsely claims the magnetic force has a component along velocity, which violates the cross-product geometry. Your study tip: always ask "what direction is the force, and what direction is the displacement?" Work requires a parallel component — perpendicularity always means zero work, regardless of force magnitude.