Question

The uniform magnetic field perpendicular to the plane of a conducting ring of radius $a$ changes at the rate of $\alpha$, then

• A. all the points on the ring are at the same potential
• B. the emf induced in the ring is $\pi a^2\alpha$
• C. electric field intensity $E$ at any point on the ring is zero
• D. $E=\left(a\alpha \right)/2$
• E. answer required

Answer 1

Answer: (A), (B), (D)

the emf induced in the ring is $\pi a^2\alpha$
$E=\left(a\alpha \right)/2$
all the points on the ring are at the same potential

$\frac{dB}{dt}=\alpha$

Thus emf induced in the ring=$\frac{d\varphi }{dt}=\frac{d\left(BA\right)}{dt}=\pi a^2\frac{dB}{dt}=\pi a^2\alpha$

From Maxwell's equation we know

$curl\left(\overrightarrow{E}\right)=-\frac{dB}{dt}$

$⟹E.2\pi a=\alpha .\pi a^2$

$⟹E=\frac{a\alpha }{2}$