Question

A long solenoid of initial radius $R_0$ is put in a region of uniform magnetic field $\mathbf{B}$ with the axis of the solenoid aligned along the magnetic field. The solenoid is a part of a closed circuit that has no initial current running through it. If the radius of the solenoid starts increasing at a uniform rate, how do the magnetic field strength $B_{in}$ and the associated magnetic energy $U_{in}$ inside the solenoid change?

• A. $B_{in}$ decreases, $U_{in}$ decreases.
• B. $B_{in}$ increases, $U_{in}$ decreases.
• C. $B_{in}$ increases, $U_{in}$ increases.
• D. $B_{in}$ decreases, $U_{in}$ increases.

Hint:

For superconducting or low-resistance loops, the total magnetic flux $\Phi$ is conserved.
Using the relation $U \propto \frac{\Phi^2}{A}$ immediately shows that as the area $A$ increases, the stored magnetic energy $U$ must decrease.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a closed loop with low resistance, the total magnetic flux must remain constant to oppose any external change (Lenz's Law Flux Conservation).

Step 2: Detailed Explanation:
1. Flux Conservation: Total flux \( \Phi = B_{\text{in}} \times \text{Area} = B_{\text{in}} \times \pi R^{2} = \text{constant} \).
As the radius \( R \) increases, the area increases. To keep \( \Phi \) constant, the magnetic field \( B_{\text{in}} \) must decrease.
2. Energy Analysis: Magnetic energy \( U_{\text{in}} = \frac{B_{\text{in}}^{2}}{2\mu_{0}} \times \text{Volume} \).
Volume of solenoid \( V = \text{Area} \times \text{length} = A \cdot \ell \).
Since \( B_{\text{in}} \propto 1/A \), then:
\[ U_{\text{in}} \propto \left(\frac{1}{A}\right)^{2} \times (A \cdot \ell) = \frac{1}{A} \]
As the area \( A \) increases, the total magnetic energy \( U_{\text{in}} \) decreases.

Step 3: Final Answer:
Both \( B_{\text{in}} \) and \( U_{\text{in}} \) decrease as the radius increases.