Question

A metal conductor of length 1 m rotates vertically about one of its ends with an angular velocity 5 rad s${}^{-1}$. If the horizontal component of earths magnetic field is 0.2 x 10${}^{-4}$ T, then the em (developed between the ends of the conductor is

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

Answer 1

Answer: (C)

$50\mu V$

The emf developed between the ends of the conductor is
$ϵ=\frac{1}{2}\omega B{l}^2$

$=\frac{1}{2}\times 5\times 0.2\times {10}^{-4}\times (1{)}^2$
$=5\times {10}^{-5}V=50\times {10}^{-6}V$

$=50\mu V$