Gujarati
5.Magnetism and Matter
medium

A bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4×10^{-5}\, T$ to a direction $ 60° $ from the field will be.....$ J$

A

$0.2$

B

$2$

C

$4.18$

D

$2 × 10^2$

Solution

(a)Magnetic moment of bar $M = {10^4}\,J/T$
$B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 M}}{{{d^3}}}$
Hence work done $W = \overrightarrow M .\overrightarrow B $
$ = {10^4} \times 4 \times {10^{ – 5}} \times \cos {60^o} = 0.2\;J$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.