Question

A bar magnet of magnetic moment M is placed in the magnetic field B. The torque acting on the magnet is:
(A)$M \times B$
(B) $M - B$
(C) \[\dfrac{1}{2}M \times B\]
(D) \[M + B\]

Answer 1

Hint: When a bar magnet is placed in a uniform magnetic field a couple force acts on the magnet which will produce torque but no translation motion. The force is not along a common line of action which leads to torque but no net force.
Formula used
$\tau = Fd$ (Where \[\tau \] is the torque and $F$is the force and $d$is the perpendicular distance between the two couple forces.)
$F = ilB$(Where $i$ is current, $B$ is magnetic field, \[l\] is the length)

Complete step by step solution:
A couple force consists of two parallel forces that are equal in magnitude, opposite in direction, and do not have the same line of action.
There’s a couple force acting on the magnet of magnetic moment (M) placed in a magnetic field (B) which in turn produces torque.
The torque produced by couple force is given by,
If \[\tau \] is the torque and $F$is the magnitude of couple force and $d$is the perpendicular distance between the two couple force then,
$\tau = Fd$
We know that if $i$ is current, $B$ is magnetic field, \[l\] is the length,
$F = ilB$
From using the two formulas stated above we can also state that,
If \[b\]is the length of magnet/wire and \[\sin \theta \]is the sin of the angle between magnet/wire and couple force then,
\[\tau = ilBb\sin \theta \]
The term $ilb$ is the dipole moment (M).
Hence, we can assert that,
\[\tau = MB\sin \theta \]
\[ \Rightarrow \tau = M \times B\]

Therefore, the answer to our question is (A)$M \times B$.

Note:
Couple forces only produce rotational motion, not translation motion. As their net force is zero but the net torque due to them is not zero. Therefore here also magnetic force will apply a net torque on the magnet and the net force will be zero.