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Question #43016

Difficulty: 8/10 | 🟢 Free

The Question

A long solenoid of length $L$ has a total of $N$ closely and uniformly wound turns. It carries a steady current $I$ . The cross-section of the solenoid is circular, but its radius varies linearly from $r_1$ at one end ( $x=0$ ) to $r_2$ at the other end ( $x=L$ ). Assume that $r_1, r_2 \ll L$ , such that the magnetic field inside the solenoid at any cross-section can be approximated by the field of an ideal infinite solenoid. What is the total magnetic energy stored inside the solenoid?

Options

A.
$\frac{\pi \mu_0 N^2 I^2}{6 L} (r_1^2 + r_1 r_2 + r_2^2)$
B.
$\frac{\pi \mu_0 N^2 I^2}{8 L} (r_1 + r_2)^2$
C.
$\frac{\pi \mu_0 N^2 I^2}{4 L} (r_1^2 + r_2^2)$
D.
$\frac{\pi \mu_0 N^2 I^2}{2 L} (r_2 - r_1)^2$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Ampere’s Circuital Law Symmetry|Energy Stored in Inductor|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply the concept of magnetic energy density over a varying volume, rather than blindly using the lumped parameter formula $U = \frac{1}{2}LI^2$ with an incorrectly averaged inductance.

A common trap is to assume the effective area is the area of the average radius $A_{eff} = \pi(\frac{r_1+r_2}{2})^2$ , or the average of the end areas.

By forcing the integration of the local energy density $u = \frac{B^2}{2\mu_0}$ over the differential volume element $dV = A(x)dx$ , the problem filters out candidates who memorize standard formulas without understanding the underlying spatial distribution of the magnetic field energy.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Ampere’s Circuital Law Symmetry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free