A conducting rod PQ of mass ‘m’ and of length ‘l’ is pl
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A conducting rod PQ of mass ‘m’ and of length ‘l’ is placed on two long parallel(smooth and conducting) rails connected to a capacitor as shown below. The rod PQ is connected to a non-conducting 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 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.
A conducting rod PQ of mass ‘m’ and of length ‘l’ is placed on two long parallel(smooth and conducting) rails connected to a capacitor as shown below. The rod PQ is connected to a non-conducting 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 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.
A. kxm
B. kxm+B2l2c
C. kxm−B2l2c
D. None of these
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Let the velocity of rod be v0 when it has been displaced by ‘x’ due to motion of rod an emf, will be induced in rod given by e = BLv0, due to this induced emf, charging of the capacitor takes place as a current, flows in the circuit [for very small time] as a result of this current, the rod experiences a magnetic force given by IBL.
From Newton's 2nd law,
LIB+Kx=ma
⇒I=ddt(Q)=ddt(C×BvL)=CBL×dvdt
⇒a=Kxm−B2L2C=w2x
