Question

A metal conductor of length $1m$ rotates vertically about one of its ends at angular velocity $5$ radians per second. If the horizontal component of the earth's magnetic field is $0.2\times {10}^{-4}T$, then the e.m.f developed between the two ends of the conductor is

• A. $5\mu V$
• B. $50\mu V$
• C. $5mV$
• D. $50mV$

Answer 1

The correct option is B

$50\mu V$

Step 1. Given data:

Length $\left(l\right)$ = $1m$

Angular velocity $\left(\omega \right)$= $5$ radians per second

Horizontal component of the earth's magnetic field $\left(B\right)$ = $0.2\times {10}^{-4}T$

Step 2. Formula used:

Induced emf $\epsilon =B\omega l^2/2$

Step 3. Calculations:

Putting all given values, we get

$$\begin{gathered} \epsilon =\left(0.2\times {10}^{-4}\times 5\times 1\times 1\right)/2 \\ =50\mu V \end{gathered}$$

Thus, the correct option is option B.