A magnetic dipole of magnetic moment $M=0.5{\text{ A m}}^2$ is placed in a magnetic field $B=0.4\text{ T}$. The work done $W$ in rotating the dipole from ${\theta }_1=0^{\circ }$ to ${\theta }_2=90^{\circ }$ is:

Solution

1. Identify the given values: magnetic moment $M=0.5{\text{ A m}}^2$, magnetic field $B=0.4\text{ T}$, initial angle ${\theta }_1=0^{\circ }$, and final angle ${\theta }_2=90^{\circ }$.
2. Use the formula for work done in rotating a magnetic dipole: $W=MB(\cos {\theta }_1-\cos {\theta }_2)$.
3. Substitute the given values into the equation: $W=(0.5)(0.4)(\cos 0^{\circ }-\cos 90^{\circ })$.
4. Calculate the result: $$\begin{array}{rl}W& =0.2\times (1-0)\\ & =0.2\text{ J}\end{array}$$.
5. Hence the answer is (B).