Question

There is a horizontal cylindrical uniform but time-varying magnetic field increasing at a constant rate $\frac{dB}{dt}$ 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. $\frac{1}{K}\left[mg-\frac{q{R}^2}{2l}\frac{dB}{dt}\right]$
• B. $\frac{1}{K}\left[mg+\frac{q{R}^2}{l}\frac{dB}{dt}\right]$
• C. $\frac{1}{K}\left[mg+\frac{2q{R}^2}{l}\frac{dB}{dt}\right]$
• D. $\frac{1}{K}\left[mg+\frac{q{R}^2}{2l}\frac{dB}{dt}\right]$

Answer 1

Answer: (D)

The electric field due to changing magnetic field will be
$E2\pi l=\pi .R^2\left(\frac{dB}{dt}\right)$
$E=\frac{R^2}{2l}\left(\frac{dB}{dt}\right)$
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 $E$ ,
$x=\frac{q.R^2}{K2l}\left(\frac{dB}{dt}\right)+\frac{mg}{K}$
$x=\frac{1}{K}\left[mg+\frac{q{R}^2}{2l}\left(\frac{dB}{dt}\right)\right]$ .