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Question: A bar magnet of magnetic moment \(M\) is placed at right angles to a magnetic induction \(B\). If a ...

A bar magnet of magnetic moment MM is placed at right angles to a magnetic induction BB. If a force FF is experienced by each pole of the magnet, the length of the magnet will be
A. MBF\dfrac{{MB}}{F}
B. BFM\dfrac{{BF}}{M}
C. MFB\dfrac{{MF}}{B}
D. FMB\dfrac{F}{{MB}}

Explanation

Solution

The question states that a bar magnet is placed at right angles to a magnetic induction. So, we shall use the formula of torque experienced by a bar magnet when it interacts with the magnetic induction which is given as τ=M×B\vec \tau = \vec M \times \vec B where τ\tau is the torque, MM is the magnetic moment and BB is magnetic induction.Also, we will use the relation that expresses torque in terms of the force experienced by a body which is given as τ=r×F\vec \tau = \vec r \times \vec F where FF is the force experienced by each pole of the magnet and rr is the distance from each pole from the point of rotation.We will then use the fact that the magnet and magnetic induction are placed at right angles. So, θ=900\theta = {90^0} .Then we shall equate both the relations to get the answer.

Complete step by step answer:
The torque on a bar magnet with a given magnetic moment when placed in a magnetic induction is given as τ=M×B\vec \tau = \vec M \times \vec B where τ\tau is the torque, MM is the magnetic moment and B is magnetic induction.This can be rewritten as,
τ=MBsinθ\tau = MB\sin \theta
Now it is given that θ=900\theta = {90^0}
τ=MBsin900\Rightarrow \tau = MB\sin {90^0}
τ=MB.......(1)\Rightarrow \tau = MB\,\,\,\,\,\,\,\,\,\,.......(1)

Now, the basic definition of torque says that it is the cross product of the force experienced at the point in consideration and the distance from the axis of rotation.
τ=r×F\vec \tau = \vec r \times \vec F
where FF is the force experienced by each pole of the magnet and rr is the distance from each pole from the point of rotation.
Hence, r=Lr = L
where LL is the length of the bar magnet.
τ=L×F\vec \tau = \vec L \times \vec F

This can be rewritten as τ=LFsinθ\tau = LF\sin \theta
Now it is given that θ=900\theta = {90^0}
τ=LFsin900\Rightarrow \tau = LF\sin {90^0}
τ=LF.......(2)\Rightarrow \tau = LF\,\,\,\,\,\,\,\,\,\,.......(2)
From (1) and (2) we get,
LF=MBLF = MB
L=MBF\therefore L = \dfrac{{MB}}{F}

Hence, option A is the correct answer.

Note: While dealing with cross products, the direction of every physical quantity in a particular physical relation must be known or derived. In this question, the direction of magnetic moment and magnetic induction was perpendicular. And since, the cross product always gives a perpendicular vector, all the three quantities-torque, magnetic moment and magnetic were mutually perpendicular. This makes it a special case and can be used directly in many numericals based on the same concept.