Question

There is a horizontal cylindrical uniform but time-varying magnetic field increasing at a constant rate $\frac{d B}{d t}$ as shown. A charged particle having charge $q$ and mass $m$ is kept in equilibrium, at the top of a spring of spring constant $K$ in such a way that it is on the horizontal line passing through the centre of the magnetic field as shown in the figure. The compression in the spring will be

• A. $ \, \frac{1}{K}\left[m g - \frac{q R^{2}}{2 l} \frac{d B}{d t}\right]$
• B. $\frac{1}{K}\left[m g + \frac{q R^{2}}{l} \frac{d B}{d t}\right]$
• C. $ \, \frac{1}{K}\left[m g + \frac{2 q R^{2}}{l} \frac{d B}{d t}\right]$
• D. $ \, \frac{1}{K}\left[m g + \frac{q R^{2}}{2 l} \frac{d B}{d t}\right]$

Answer 1

Answer: (D)

The electric field due to changing magnetic field will be
$E2\pi l=\pi .R^{2}\left(\frac{d B}{d t}\right)$
$E=\frac{R^{2}}{2 l}\left(\frac{d B}{d t}\right)$
Using Lenz law we can find that the direction of this magnetic field is downward at the charge. Therefore, total downward force that the spring balances is,
$qE+mg=Kx$
Putting the value of $E$ ,
$x=\frac{q . R^{2}}{K 2 l}\left(\frac{d B}{d t}\right)+\frac{m g}{K}$
$x=\frac{1}{K}\left[m g + \frac{q R^{2}}{2 l} \left(\frac{d B}{d t}\right)\right]$ .