Question

A simple pendulum with bob of mass $m$ and conducting wire of length $L$ swings under gravity through an angle $\theta$. The component of earth's magnetic field, in a direction perpendicular to swing is $B$. Then

• A. The maximum velocity of pendulum is $2\sqrt{gL}\sin \left(\frac{\theta }{2}\right)$
• B. The maximum velocity of pendulum is $\sqrt{gL}\sin \left(\frac{\theta }{2}\right)$
• C. The maximum potential difference induced across the pendulum is $BL\sqrt{gL}\sin \left(\frac{\theta }{2}\right)$
• D. The maximum potential difference induced across the pendulum is $2BL\sqrt{gL}\sin \left(\frac{\theta }{2}\right)$

Answer 1

Answer: (A), (D)

The maximum potential difference induced across the pendulum is $2BL\sqrt{gL}\sin \left(\frac{\theta }{2}\right)$


From the figure,

$h=L(1-\cos \theta )$

The maximum emf will be induced when the velocity of the pendulum is maximum, i.e. when the pendulum is at the lowest position.

The maximum velocity of the pendulum is,

$v_{\text{max}}^2=u^2+2gh$

$=0^2+2gL(1-\cos \theta )[\because u=0$ ]

$=2gL\times 2{\sin }^2\left(\frac{\theta }{2}\right)$

$\therefore v_{\text{max}}=2\sqrt{gL}\sin \left(\frac{\theta }{2}\right)$

So, option $\left(\text{A}\right)$ is correct.

Now, the maximum induced potential difference is,

$E_{max}=B{v}_{max}L$

$=B\times 2\sqrt{gL}\sin \left(\frac{\theta }{2}\right)\times L$

$\therefore E_{max}=2BL\sqrt{gL}\sin \left(\frac{\theta }{2}\right)$

Thus, option $\left(\text{D}\right)$ is also correct.