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Question: A short bar magnet placed in a uniform magnetic field experiences a maximum couple of \[4 \times {10...

A short bar magnet placed in a uniform magnetic field experiences a maximum couple of 4×10−5N−m4 \times {10^{ - 5}}N - m. The magnet is turned through 45∘45^\circ . The couple acting in the new equilibrium position is
A) 2×10−5 Nm2 \times {10^{ - 5}}\,Nm
B) 2.828×10−5 Nm2.828 \times {10^{ - 5}}\,Nm
C) 1.414×10−5 Nm1.414 \times {10^{ - 5}}\,Nm
D) 3.6×10−5 Nm3.6 \times {10^{ - 5}}\,Nm

Explanation

Solution

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
τ=m×B\tau = m \times B where τ\tau is the magnetic couple acting on an object with a magnetic moment mm when placed in a magnetic field of strength BB

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
τ=m×B\tau = m \times B
The cross product in the above term can be simplified as
τ=mBsin⁡θ\tau = mB\sin \theta
The maximum couple experienced by the magnet is 4×10−5N−m4 \times {10^{ - 5}}N - m which corresponds to θ=90∘ (as sin90∘ = 1)\theta = 90^\circ \,({\text{as}}\,{\text{sin90}}^\circ {\text{ = 1)}}. Thus we can write the maximum couple as
τmax=mB=4×10−5{\tau _{max}} = mB = 4 \times {10^{ - 5}}
When the magnet is rotated through 45∘45^\circ , the new coupling will be calculated as
τ=mBsin⁡45∘\tau = mB\sin 45^\circ
⇒τ=mB12\Rightarrow \tau = mB\dfrac{1}{{\sqrt 2 }}
Hence the magnetic couple will be
τ=4×10−5×12\tau = 4 \times {10^{ - 5}} \times \dfrac{1}{{\sqrt 2 }}
Which gives us

τ=2.82×10−5 N/m\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 θ=0∘\theta = 0^\circ . Hence we would have to hold the magnet in its place to form an angle of 45∘ or 90∘45^\circ \,{\text{or}}\,90^\circ as otherwise, it will just align with the external magnetic field.