Question

A straight conductor of length 0.6 m is moved with a speed of 10 $\text{ms}^{-1}$ perpendicular to a magnetic field of induction 1.2 $\text{Wb/m}^2$. The induced e.m.f. across the conductor is

• A. 6 V
• B. 7.2 V
• C. 0.72 V
• D. 12 V

Hint:

Multiplying the speed $10$ by the decimal length $0.6$ simplifies things to a clean whole number $6$ instantly. Then, $6 \times 1.2 = 7.2\ \text{V}$ becomes a simple mental calculation!

Answer 1

The Correct Option is B

When a straight conducting rod moves perpendicularly through a uniform magnetic field, the motional electromotive force ($e$) induced across its ends is given by: $$e = B l v$$ Given parameters: * Magnetic induction, $B = 1.2\ \text{Wb/m}^2$ * Length of conductor, $l = 0.6\ \text{m}$ * Velocity, $v = 10\ \text{ms}^{-1}$ Substituting these values into the equation: $$e = 1.2 \times 0.6 \times 10 = 1.2 \times 6 = 7.2\ \text{V}$$
Final Answer:
The induced e.m.f. across the conductor is 7.2 V, matching option (B).