Q.

A conducting rod PQ of mass m and length l is placed on two long parallel (smooth and conducting) rails connected to a capacitor as shown. The rod PQ is connected to a non conducting spring of spring constant k, which is initially in relaxed state. The entire arrangement is placed in a magnetic field perpendicular to the plane of figure. Neglect the resistance of the rails and rod. Now, the rod is imparted a velocity v0 towards right, then acceleration of the rod as a function of its displacement x is given by

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a

kxmB2/2C

b

kxm+B2I2C

c

kxB2I2c

d

kxm

answer is C.

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Detailed Solution

Let the velocity of rod be v when it has been displaced by x . Due to motion of rod an emf will be induced in rod given by e = Bvl , due to this induced emf, charging of the capacitor takes place  as a current, flows in the circuit [for very small time] a result of this current, the rod experiences a magnetic force given by IBl .

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From Newton’s second law,
BI+kx=maa=dvdtI=ddt[Q]=ddt[C×BvI]=CBI×dvdta=kxmB2I2C=ω2x
Which also shows that rod is performing SHM

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