Question

A rod of $20cm$ length and mass $250g$ is connected to a two spring system as shown in the Figure. The other end of each spring is attached to a fixed support. A uniform magnetic field of $0.4T$ is directed perpendicular and into the plane of the spring-rod system. At $t=0$, the rod is released with the springs extended by $20cm$. If the spring constant of each spring is $2N{m}^{-1}$, then,

• A. the maximum value of emf induced across the rod is $64mV$.
• B. the induced emf across the rod is reduced to zero in $\frac{\pi }{8}$ S after releasing the rod.
• C. the induced emf across the rod reaches maximum first time in $\frac{\pi }{8}$ s after releasing the rod.
• D. the maximum value of induced emf is $32mV$.

Answer 1

Answer: (A), (C)

The correct options are
A the maximum value of emf induced across the rod is $64mV$.
C the induced emf across the rod reaches maximum first time in $\frac{\pi }{8}$ s after releasing the rod.

For parallel spring-rod system (executing SHM), angular velocity $\omega =\sqrt{\frac{2k}{m}}=4rad/s$

Time period $T=\frac{2\pi }{\omega }=\frac{\pi }{2}sec$

It is the same time in which the spring reaches to its natural length and thus rod has maximum velocity at this point and cuts the maximum flux so the rate of change of flux is maximum and thus maximum value of emf induced at this point.

Let $v_o$ be the maximum velocity of the rod when it crosses the equillibrium point.

$\frac{1}{2}m{v_o}^2=\frac{1}{2}K{x_{max}}^2$

$v_o=\omega \times x_{max}=0.80m/s$

Maximum value of emf induced $e_{max}=Bl{v}_o=64mV$ which is at time $t=\frac{\pi }{8}sec$

Hence option (a),(c) are correct