Question

A uniform magnetic field b = kt, where 'k' is a constant. Is applied perpendicular to the plane of a non- conducting loop having a charge q, mass'm'. The angular velocity of the loop at time 't' is:

Answer 1

Given that,

Magnetic field is

$b=kt$

k = constant

Charge = q

Mass = m

Consider a ring of radius r, width dr and charge on ring

The charge on the ring is

$dq=(2\pi rdr)-\dfrac{Q}{\pi R^2}$

Induced emf in ring

$\epsilon=\int{\vec{E}\cdot \vec{dl}}=-\dfrac{-d}{dt}\int{\vec{B}\cdot \vec{dS}}$

We need to calculate the electric field in circular loop

Using formula of electric field

$E 2\pi=-\pi r^2\dfrac{db}{dt}$

$E=-\dfrac{r.k}{2}$

We need to calculate the force on the ring

Using formula of force

$dF=dq E$

$dF=-(dq)\dfrac{rk}{2}$

We need to calculate the value of α

Using formula of torque

$\tau=\int_{0}^{R}{rdF}$

$\tau=\int_{0}^{R}{r(\dfrac{2\pi Qdr}{\pi T^2})\dfrac{rk}{2}}$

$\tau=\int_{0}^{R}{\dfrac{kQr^3}{R^2}}dr$

$\tau=\dfrac{-kQR^2}{4}$

$I\alpha=\dfrac{-kQR^2}{4}$

$\alpha=-\dfrac{kQR^2}{4(\dfrac{mR^2}{2})}$

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

We need to calculate the angular velocity of the loop

Using formula of angular velocity

$\omega=\alpha t$

Put the value into the formula

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

The magnitude of angular velocity

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

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