Question

A conductor of length $5\,cm $ is moved parallel to itself with a speed of $2\,m/s, $ perpendicular to a uniform magnetic field of $10^{-3} Wb/m^{2}$ . The induced e.m.f. generated is

• A. $ 2\times 10^{-3}V$
• B. $ 1\times 10^{-3}V $
• C. $ 1\times 10^{-4}V $
• D. $ 2\times 10^{-4}V$

Answer 1

Answer: (C)

$e=-B v l$
Given, $l=5\, cm =5 \times 10^{-2} m,\, v=2\, m /s$
$B=10^{-3} Wb / m ^{2}$
$e=10^{-3} \times 2 \times 5 \times 10^{-2}$
$e=10 \times 10^{-5} V$
$e=1 \times 10^{-4} V$