Question

A bar magnet whose magnetic dipole moment is 21 A·m2 is aligned with an applied magnetic field of 3.6 T. How much work must you do to rotate the bar magnet 180° to point in the direction opposite to the magnetic field?

Answer 1

Answer:

The amount of work must be do to rotate the bar magnet is 151.2 J

Explanation:

Given:

Magnetic moment $\mu = 21$ $A. m^{2}$

Magnetic field $B = 3.6$ T

To find work do to rotate the bar magnet,

From the formula of work done in case of magnetic field,

    $U = \mu .B \cos 0 -\mu .B \cos 180$

Here $\theta$ changes 0 to 180

But $\cos 180 = -1$

    $U = 2\mu B$

    $U = 2 \times 21 \times 3.6$

    $U = 151.2$ J

Therefore, the amount of work must be do to rotate the bar magnet is 151.2 J