Question

A bar magnet whose magnetic dipole moment is $\mathbf{15}{\mathbf{\text{\hspace{0.17em}A.m}}}^{\mathbf{\text{2}}}$is aligned with an applied magnetic field of $\mathbf{4}\mathbf{\text{\hspace{0.17em}T}}$. How much work must you do to rotate the bar magnet $\mathbf{180}\mathbf{°}$to point in the direction opposite to the magnetic field?

Short answer

The work done to rotate the bar magnet $180°$to point in the direction opposite to the magnetic field is 120 J.

Step-by-step solution

Given Information

The dipole moment of the magnet is $\mu =15{\text{\hspace{0.17em}A.m}}^{\text{2}}$, and the magnetic field is $B=4\text{\hspace{0.17em}T}$. T. The bar rotates at an angle $\theta =180°$.

Work done

The directions of the bar rotation and the magnetic torque are the same, therefore, the work done is non-zero and equals to the difference between the two potential energies at the angle ${\theta }_1=0$and ${\theta }_2=180°$given as:

$W=\Delta U_2-\Delta U_1\cdots \left(1\right)$

Where the potential energy for a magnetic dipole is given by

$\begin{array}{c}U=-\overrightarrow{\mu }\cdot \overrightarrow{B}\\ =-\mu B\cos \theta \end{array}$

Therefore, the equation can be written as:

$\begin{array}{c}W=\Delta U_1-\Delta U_2\\ =-\mu B\cos {\theta }_2-\left(-\mu B\cos {\theta }_1\right)\\ =\mu B\left(\cos {\theta }_1-\cos {\theta }_2\right)\cdots \left(2\right)\end{array}$

Calculate the work done

Substitute the given values of $\mu ,B,{\theta }_1$and ${\theta }_2$into equation (2) to get the work done.

$\begin{array}{c}W=\mu B\left(\cos {\theta }_1-\cos {\theta }_2\right)\\ =15\cdot 4\left(\cos 0°-\cos 180°\right)\\ =60\left(1-\left(-1\right)\right)\\ =120\text{\hspace{0.17em}J}\end{array}$

Thus, the work done is 120 J.