Question

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 1

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 $\dfrac{d\tau }{d\theta }$ 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=$\dfrac{d\tau }{d\theta }$
$\dfrac{d\tau }{d\theta }=\dfrac{d(MB\sin \theta )}{d\theta }$
$\dfrac{d\tau }{d\theta }=MB\cos \theta $
For $\dfrac{d\tau }{d\theta }$ 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.