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Question #43422

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The Question

A non-magnetic toroid has a rectangular cross-section with inner radius $a$ , outer radius $2a$ , and height $h$ . The toroid is centered at the origin with its axis of symmetry along the $z$ -axis. A primary coil of $N_1$ turns is uniformly wound on the toroid and carries a time-varying current $I(t) = I_0 \frac{t}{t_0}$ . The primary coil is wound such that the magnetic field it generates inside the toroid is in the $+\hat{\theta}$ direction.

The entire setup is permeated by an external time-varying magnetic field given in cylindrical coordinates $(r, \theta, z)$ as: $\vec{B}_{ext}(r, t) = B_0 \frac{t}{t_0} \left( \frac{a}{r} \hat{\theta} + \frac{r}{a} \hat{k} \right)$

A secondary coil of $N_2$ turns is wound purely poloidally (no net azimuthal winding) around the toroid's cross-section. The total resistance of the secondary coil is $R$ . Assuming the magnetic field generated by the induced current in the secondary coil is negligible, which of the following expressions represents the total thermal energy dissipated in the secondary coil from $t=0$ to $t=t_0$ ?

Options

A.
$\frac{N_2^2 h^2 \ln^2(2)}{R t_0} \left( \frac{\mu_0 N_1 I_0}{2\pi} + B_0 a \right)^2$
B.
$\frac{N_2^2 h^2 \ln^2(2)}{R t_0} \left( \frac{\mu_0 N_1 I_0}{2\pi} \right)^2$
C.
$\frac{N_2^2 h^2}{R t_0} \left( \frac{\mu_0 N_1 I_0}{2\pi} \ln(2) + \frac{3}{2} B_0 a \right)^2$
D.
$\frac{N_2^2 h^2}{R t_0} \left( \frac{\mu_0 N_1 I_0}{2\pi} \ln(2) + B_0 a \ln(2) + \frac{3}{2} B_0 a \right)^2$

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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

Toroidal Solenoid Boundary Conditions|Mutual Inductance Reciprocity|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply Faraday's Law and the superposition of magnetic flux in a toroidal geometry.

The primary trap lies in the vector alignment of the external magnetic field.

Students often mechanically calculate the self-inductance flux of the toroid while either ignoring the external field entirely or incorrectly integrating its magnitude over the cross-section.

By providing an external field with both azimuthal ( $\hat{\theta}$ ) and axial ( $\hat{k}$ ) components, the problem forces the student to recognize that only the azimuthal component penetrates the poloidal area vector ( $d\vec{A} = h \, dr \, \hat{\theta}$ ).

Furthermore, it connects the macroscopic Joule heating to the Work-Energy Theorem, reinforcing that the thermal energy is the work done by the non-conservative induced electric field.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Toroidal Solenoid Boundary Conditions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free