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 Question:
We have a closed-loop conducting solenoid placed in an external magnetic field $\mathbf{B}$ parallel to its axis.
The radius of the solenoid is increasing.
As the cross-sectional area of the solenoid increases, the magnetic flux through it changes, inducing an electromotive force (EMF) and a current to oppose this change.

Step 2: Key Formula or Approach:
According to Lenz's law and flux conservation in a low-resistance closed circuit, the total magnetic flux $\Phi$ through the solenoid is conserved:
\[ \Phi = B_{in} A = \text{constant} \] The magnetic energy $U_{in}$ inside the solenoid is given by:
\[ U_{in} = \frac{B_{in}^2}{2\mu_0} \times \text{Volume} \]

Step 3: Detailed Explanation:

• Let the cross-sectional area of the solenoid be $A = \pi R^2$.

• As the radius $R$ increases, the area $A$ increases.

• To maintain flux conservation, we have:
\[ B_{in} \cdot A = \text{constant} \implies B_{in} \propto \frac{1}{A} \] Since $A$ is increasing, $B_{in}$ must decrease.

• The magnetic energy density is:
\[ u = \frac{B_{in}^2}{2\mu_0} \]
• The total magnetic energy $U_{in}$ inside a solenoid of length $l$ is:
\[ U_{in} = u \cdot \text{Volume} = \frac{B_{in}^2}{2\mu_0} (A \cdot l) \]
• Since $B_{in} \propto \frac{1}{A}$:
\[ U_{in} \propto \left( \frac{1}{A} \right)^2 \cdot A = \frac{1}{A} \]
• Since the area $A$ increases, the magnetic energy $U_{in}$ must decrease.

Step 4: Final Answer:
Both $B_{in}$ and $U_{in}$ decrease as the radius increases.