Question

A metal conductor of length $1$m rotates vertically about one of its ends at angular velocity $5$ radians per second. If the horizontal component of earth's magnetic field is $0.3\times {10}^{-4}T$, then the e.m.f developed between the two ends of the conductor is

• A. $100\mu V$
• B. $50\mu V$
• C. $150\mu V$
• D. $75\mu V$

Answer 1

The correct option is D $75\mu V$

$Given:$ Length = 1m, w=5rad/s, B=$0.3$ $\times$$10$${}^{-}4$ $T$

$Solution:$ We know that,

$e$ $=$ $\frac{1}{2}$ $B$$\omega$$L^2$

=$\frac{1}{2}\times 0.3\times {10}^{-4}$ x5 x (1)${}^2$

$=$ $7.5$ $\times$ ${10}^{-5}$

$=$ $75\mu V$