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

Step 1: Recall the motional emf.
A rod of length $l$ moving with speed $v$ across a field $B$ produces emf $e=Blv$.

Step 2: Put in the values.
$B=1.2$, $l=0.6$, $v=10$. \[ e=1.2\times0.6\times10 \]

Step 3: Multiply.
$e=7.2$ V. \[ \boxed{7.2\text{ V}} \]