Question

A circular ring of diameter \(20\) cm has a resistance of \(0.01\) \(\Omega\). The charge that will flow through the ring, if it is turned from a position perpendicular to a uniform magnetic field of \(2\)T to a position parallel to the field is (in coulomb) (\(\because \cos 0^{\circ }=1\), \(\cos 90^{\circ }=0\))

• A. \(\pi\)
• B. \(2\pi\)
• C. \(\pi /2\)
• D. \(4\pi\)

Hint:

Charge equals the change in flux divided by resistance.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The induced charge is \(q=\frac{\Delta\Phi}{R}\).

Step 2: Find the Flux Change
Radius \(r=0.1\) m, so area \(=\pi r^{2}=0.01\pi\) m\(^2\). Initially \(\Phi_i=BA\cos 0^{\circ}=2\times 0.01\pi\), finally \(\Phi_f=BA\cos 90^{\circ}=0\).

Step 3: Magnitude of Change
\(|\Delta\Phi|=0.02\pi\) Wb.

Step 4: Calculation
\(q=\frac{0.02\pi}{0.01}=2\pi\) C, option (B).

Final Answer:
The charge that flows is 2 pi coulomb. \[ \boxed{\text{(B) }2\pi} \]