Question

A straight line conductor of length 0.4 m is moved with a speed of \(7.0\text{ ms}^{-1}\) perpendicular to magnetic field of intensity \(0.8\text{ Wb m}^{-2}\). The induced e.m.f. across the conductor is

• A. \(2\cdot 24\) V
• B. \(2\cdot 80\) V
• C. \(3\cdot 20\) V
• D. \(5\cdot 60\) V

Hint:

Self-inductance goes as N squared times area over length.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
\(L = \frac{\mu_0N^2A}{l} = \frac{\mu_0N^2\pi r^2}{l}\). With equal turns, \(L\propto\frac{r^2}{l}\).

Step 2: Ratio:
The lengths and radii of A and B are in the ratio \(1:3\). So \(\frac{L_A}{L_B} = \frac{r_A^2/l_A}{r_B^2/l_B} = \frac{1/1}{9/3} = \frac13\).

Final Answer:
The ratio of self-inductances is \(1 : 3\), option (B). \[ \boxed{1 : 3} \]