Question

A uniform magnetic field $B=kt$, where ${}^{\prime }{k}^{\prime }$ is a constant. is applied perpendicular to non-conducting loop having a charge $Q$, mass ${}^{\prime }{m}^{\prime }$. The angular velocity of the loop at time $t$ is:

• A. zero
• B. $\frac{kQt}{2m}$
• C. $\frac{kQ{t}^2}{4m}$
• D. Cannot be found.

Answer 1

The correct option is B $\frac{kQt}{2m}$

$Given:$

Magnetic field is

$B=kt$

$k=$ constant

Charge $=Q$

Mass $=m$

$Solution:$

Consider a ring of radius $r,$ width $dr$ and

charge on ring

The charge on the ring is

Induced emf in ring

$ϵ=\int \overrightarrow{E}\cdot \overrightarrow{d}=-\frac{-d}{dt}\int \overrightarrow{B}\cdot \overrightarrow{dS}$

We need to calculate the electric field in

circular loop

Using formula of electric field

$E2\pi =-\pi r^2\frac{db}{dt}$

$E=-\frac{r\cdot k}{2}$

We need to calculate the force on the ring

Using formula of force

$dF=dQE$

$dF=-\left(dQ\right)\frac{rk}{2}$

We need to calculate the value of $a$

Using formula of torque

$\tau ={\int }_0^{R}rdF$

$\tau ={\int }_0^{R}r\left(\frac{2\pi Qdr}{\pi T^2}\right)\frac{rk}{2}$

$\tau ={\int }_0^R\frac{kQ{r}^3}{R^2}dr$

$\tau =\frac{-kQ{R}^2}{4}$

$I\alpha =\frac{-kQ{R}^2}{4}$

$\alpha =-\frac{kQ{R}^2}{4\left(\frac{m{R}^2}{2}\right)}$

$\alpha =\frac{-kQ}{2m}$

We need to calculate the angular velocity of the loop

Using formula of angular velocity

$\omega =\alpha t$

Put the value into the formula

$\omega =\frac{-kQ}{2m}t$

The magnitude of angular velocity

$|\omega |=\frac{kQ}{2m}t$

Hence, The angular velocity of the loop is $\frac{kQ}{2m}t$

$So,the$ $correct$ $option:B$