Q113 / 150 · MHT CET 2021 (20 Sep Shift 1) · Physics

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

  • A
    $5 \mu \mathrm{V}$
  • B
    $50 \mathrm{mV}$
  • C
    $5 \mathrm{mV}$
  • Correct answer
    D
    $50 \mu \mathrm{V}$

Solution

$\begin{aligned} & \text { Induced emf, } \mathrm{e}=\frac{1}{2} \mathrm{~B} \omega \ell^2 \\ & =\frac{1}{2} \times 0.2 \times 10^{-4} \times(1)^2 \times 5 \\ & =0.5 \times 10^{-4} \mathrm{~V}=50 \mu \mathrm{V}\end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions