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The rate of change of torque τ with deflection θ is maximum for a magnet suspended freely in a uniform magnetic field of induction B when θ is equal to
A . 0°
B . 45°
C . 60°
D . 90°




Answer
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Hint:In this question we have to find the θ at which the rate of change of torque with deflection is maximum. Therefore, we have to make dτdθ maximum.





Formula used:
Torque τ= M×B (the cross product of magnetic moment and magnetic field)






Complete answer:

We know that for a magnet suspended freely in a uniform magnetic field, the torque is the cross product of the magnetic moment and the magnetic field.
τ= M×B
τ= MBsinθ (θ=angle between M and B)
The rate of change of torque τ with deflection θ=dτdθ
dτdθ=d(MBsinθ)dθ
dτdθ=MBcosθ
For dτdθ to be maximum, MBcosθ should be maximum, that is cosθ should be maximum since we cannot change the magnetic moment or the magnetic field.
The maximum value of cosθ is 1 which happens when θ=0°.
Therefore, the rate of change of torque with deflection is maximum when vector M and vector B are parallel to each other.
The correct answer is 0°.






Note:The differentiation of sin θ with respect to θ gives cosθ with a positive sign. If we have to find the rate of change of something, we always take help of differentiation.