Question

A magnet of magnetic moment \[50\widehat{i}\] \[A{{m}^{2}}\] is placed along the x- axis in a magnetic field \[\overrightarrow{B}=0.5\widehat{i}+3\widehat{j}\]. The torque acting on the magnet is?
A. 175k Nm
B. 150k Nm
C. 75k Nm
D. \[25\sqrt{37}\widehat{k}\]Nm

Answer 1

Hint: When a magnetic dipole is placed in an external magnetic field, then both the ends of the dipole experiences a force but these are force cancels out each other, but a net torque acts on the dipole. The torque acting on a dipole placed in a magnetic field can be found by using the formula torque= force times perpendicular distance between forces.

Complete step by step answer:
Given magnetic dipole moment, M= \[50\widehat{i}\] \[A{{m}^{2}}\]
Magnitude of field along with direction, \[\overrightarrow{B}=0.5\widehat{i}+3\widehat{j}\]
In vector notation, torque is given by:
\[\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}\]
$\Rightarrow \overrightarrow{\tau }=50\widehat{i}\times (0.5\widehat{i}+3\widehat{j}) \\ \Rightarrow \overrightarrow{\tau }=50\widehat{i}\times 0.5\widehat{i}+50\widehat{i}\times 3\widehat{j} \\
\Rightarrow \overrightarrow{\tau }=0+150\widehat{k} \\
\therefore \overrightarrow{\tau }=150\widehat{k}$

So, the magnitude of torque is 150 Nm and it is directed along a positive z direction, so the correct option is B.

Additional Information:
Cross product gives the result which is a vector quantity. So we have to consider the direction too. The torque acting on the dipole also leads to storage of potential energy in the dipole.

Note:While calculating torque we have to keep in our mind that it is a vector quantity and is given by the cross product of force acting on the body and the perpendicular distance from the axis of rotation. When a dipole is placed parallel to the magnetic field the net force on it is zero and there is no torque. But when it is placed at some angle with the field then it experiences net torque.