Question

A magnet having magnetic moment $\overrightarrow{M}$ = (10$\hat{i}$) Am² is placed in a magnetic field $\overrightarrow{B}$ = (2$\hat{i}$ + 3$\hat{j}$) T. The torque acting on the magnet is:

• A. 10$\hat{k}$ Nm
• B. 20$\hat{k}$ Nm
• C. 30$\hat{k}$ Nm
• D. 40$\hat{k}$ Nm

Answer 1

Answer: (C)

$30\hat{k}$ Nm

Answer 2

The torque $\overrightarrow{\tau}$ acting on a magnetic dipole with magnetic moment $\overrightarrow{M}$ placed in a magnetic field $\overrightarrow{B}$ is given by the cross product: $\overrightarrow{\tau} = \overrightarrow{M} \times \overrightarrow{B}$ Given the magnetic moment $\overrightarrow{M} = (10\hat{i})$ Am² and the magnetic field $\overrightarrow{B} = (2\hat{i} + 3\hat{j})$ T. We calculate the torque as follows: $\overrightarrow{\tau} = (10\hat{i}) \times (2\hat{i} + 3\hat{j})$ Using the distributive property of the cross product: $\overrightarrow{\tau} = (10\hat{i} \times 2\hat{i}) + (10\hat{i} \times 3\hat{j})$ Recall the properties of the cross product for unit vectors: $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$. $\overrightarrow{\tau} = (10 \times 2)(\hat{i} \times \hat{i}) + (10 \times 3)(\hat{i} \times \hat{j})$ $\overrightarrow{\tau} = 20(0) + 30(\hat{k})$ $\overrightarrow{\tau} = 0 + 30\hat{k}$ $\overrightarrow{\tau} = 30\hat{k} \text{ Nm}$