Question

A short bar magnet placed in a uniform magnetic field experiences a maximum couple of \[4 \times {10^{ - 5}}N - m\]. The magnet is turned through \[45^\circ \]. The couple acting in the new equilibrium position is
A) $2 \times {10^{ - 5}}\,Nm$
B) $2.828 \times {10^{ - 5}}\,Nm$
C) $1.414 \times {10^{ - 5}}\,Nm$
D) $3.6 \times {10^{ - 5}}\,Nm$

Answer 1

Hint: When a magnet is placed in an external field, it experiences a torque which is also known as a magnetic couple. The magnitude of this magnetic couple depends on the strength of the magnetic field, the strength of the magnet, and their orientation.
Formula used: In this solution, we will use the following formula
$\tau = m \times B$ where $\tau $ is the magnetic couple acting on an object with a magnetic moment $m$ when placed in a magnetic field of strength $B$

Complete step by step answer:
The magnetic moment is a determination of its tendency to get arranged through a magnetic field. When placed in a magnetic field, a magnet will tend to rotate such that it can align itself with the external magnetic field. The magnetic couple determines its tendency to align with the magnetic field. The couple acting on a magnet when placed in a magnetic field can be determined as
$\tau = m \times B$
The cross product in the above term can be simplified as
$\tau = mB\sin \theta $
The maximum couple experienced by the magnet is \[4 \times {10^{ - 5}}N - m\] which corresponds to $\theta = 90^\circ \,({\text{as}}\,{\text{sin90}}^\circ {\text{ = 1)}}$. Thus we can write the maximum couple as
${\tau _{max}} = mB = 4 \times {10^{ - 5}}$
When the magnet is rotated through \[45^\circ \], the new coupling will be calculated as
$\tau = mB\sin 45^\circ $
$ \Rightarrow \tau = mB\dfrac{1}{{\sqrt 2 }}$
Hence the magnetic couple will be
$\tau = 4 \times {10^{ - 5}} \times \dfrac{1}{{\sqrt 2 }}$
Which gives us

\[\tau = 2.82 \times {10^{ - 5}}\,N/m\] which corresponds to option (B).

Note: The torque experienced by the bar magnet will always be towards the equilibrium position where $\theta = 0^\circ $. Hence we would have to hold the magnet in its place to form an angle of \[45^\circ \,{\text{or}}\,90^\circ \] as otherwise, it will just align with the external magnetic field.