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Question: The time period of a freely suspended magnet does not depend upon? A. Length of magnet B. The po...

The time period of a freely suspended magnet does not depend upon?
A. Length of magnet
B. The pole strength of the magnet
C. The horizontal component of magnetic field of earth
D. The length of suspension

Explanation

Solution

The dependency of any physical quantity on other physical quantity can be checked by deriving the related formulas and checking the presence of terms in that formula. When a magnet is suspended freely in a magnetic field, then due to interaction of magnetic pole with the magnetic field, the magnet experiences a torque. For a freely suspended magnet, the horizontal component of the magnetic field is responsible for the torque.

Formula Used :
τ\tau= -mBsinθmBsin \theta, τ\tau= IαI \alpha [ I is moment of inertia about centre ]

Complete step-by-step answer :
We know, if the pole strength of a magnet is ‘m’ then in presence of a magnetic field ‘B’, it experiences a force equal to m×Bm \times B. Now suppose a magnet is freely suspended at midpoint in the earth's atmosphere, initially with length in the direction of the magnetic field. Now if the magnet is turned with an angle θ\theta, the restoring torque on it will be (note, only the horizontal component of earth’s magnetic field is responsible for the oscillations, as there’s no role of vertical component in oscillation in horizontal plane) :
τ\tau= -mBsinθmBsin \theta [m×Bm \times B=mBsinθmBsin \theta(cross product) ]
As we know, τ\tau= IαI \alpha [ I is moment of inertia about centre ]
Hence IαI \alpha= -mBsinθmBsin \theta
Now, for small angles, sinθ=θsin \theta = \theta
So IαI \alpha= -mBθmB \theta
Or α=mBIθ\alpha =- \dfrac{mB}I \theta
Now, writing α\alpha in differential form:
α\alpha= d2θdt2\dfrac{d^2 \theta}{dt^2}
Hence, d2θdt2\dfrac{d^2 \theta}{dt^2} =mBIθ=- \dfrac{mB}I \theta
Now, comparing the equation with general equation of S.H.M,
d2θdt2\dfrac{d^2 \theta}{dt^2}= - ω2θ\omega^2 \theta
We get, ω2=mBI\omega^2 = \dfrac{mB}{I}
Hence ω=mBI\omega = \sqrt { \dfrac{mB}{I}}
Now, as T = 2πω\dfrac{2\pi}{\omega}
So T=2πImBT = 2 \pi { \sqrt{ \dfrac{I}{mB}}}
Hence Time period of a freely suspended magnet does not depend upon length of suspension i.e. length of thread to which magnet is tied.
Hence D. is the correct answer.

Note :The moment of inertia (I) depends upon the length of magnet. Pole strength also plays a role in time period. Students must note that the magnetic field term appearing ‘B’ in the formula is the ‘Horizontal component of earth’s magnetic field’. Since the magnet is freely suspended, it means it can oscillate in the horizontal plane only, so nothing to do with the vertical component.