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Question #42942

Difficulty: 8/10 | 🟢 Free

The Question

A cylindrical region of radius $R$ (in $\text{m}$ ) centered on the $z$ -axis contains a uniform time-varying magnetic field $\vec{B} = \alpha t \hat{k}$ for $\sqrt{x^2+y^2} \le R$ , where $\alpha > 0$ is a constant in $\text{T/s}$ and $t$ is time in seconds. A uniform static electric field $\vec{E}_{stat} = E_0 \hat{i}$ (where $E_0 > 0$ is in $\text{V/m}$ ) is also present in this region. A point charge $+q$ (in $\text{C}$ ) is moved from point $P(R, 0, 0)$ to point $Q(-R, 0, 0)$ along the semi-circular arc $x^2 + y^2 = R^2$ ( $y \ge 0$ ). What is the total work done (in Joules) by the net electric field on the charge $q$ ?

Options

A.
$-q \left( 2 E_0 R + \frac{\pi \alpha R^2}{2} \right)$
B.
$-2 q E_0 R$
C.
$-q \left( 2 E_0 R - \frac{\pi \alpha R^2}{2} \right)$
D.
$-q \left( 2 E_0 R + \pi \alpha R^2 \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Conservative Field Path Independence|Conservative Force Field|Electromagnetic Induction - Class 12

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental distinction between conservative electrostatic fields and non-conservative induced electric fields.

A common trap for students is to assume that the total electric field is conservative and that work done is path-independent.

If a student incorrectly applies the concept of potential difference ( $\Delta V$ ) to the entire field, they will likely evaluate the work along the simplest path—the straight line from $P$ to $Q$ along the x-axis.

Along this straight path, the induced electric field $\vec{E}_{ind}$ is perpendicular to the displacement, yielding zero work for the induced component, leading them to the incorrect answer $-2qE_0R$ .

The problem forces the realization that in the presence of a time-varying magnetic field, $\oint \vec{E} \cdot d\vec{l} \neq 0$ , making the work done strictly path-dependent.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Conservative Field Path Independence
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free