- Home
- Standard 12
- Physics
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
medium
