Question

Determine the energy required to reverse the direction of a magnetic dipole of 3×10^-4 A.m² in a magmetic field of 0,6 T

Answer 1

Gives

$M=3\times10^{-4}Am^2$

$B=0.6T$

$\theta=180$

Now

Energy $U=-M.B$

$U= -MBcos\theta$

$cos180°=-1$

$U=-3\times10^{-4}\times0.6\times cos180°$

$U=1.8\times20^{-4} J$