Question

If a uniform magnetic field $B$, 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 magnitude of emf induced in the ring is $\pi a^2\alpha$
• C. electric field intensity $E$ at any point on the ring is zero
• D. the magnitude of the electric field intensity $|E|=\left(a\alpha \right)/2$

Answer 1

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

the magnitude of the electric field intensity $|E|=\left(a\alpha \right)/2$

$\varphi =\pi a^{2}B$

$|e|=\pi a^2\frac{dB}{dt}=\pi a^2\alpha$

$E2\pi a=e\Rightarrow E=\frac{\pi a^2\alpha }{2\pi a}=\frac{\alpha a}{2}$

Let $R$ be the resistance of the ring. Then current in the ring is $i=e/R$

Consider a small element $d\ell$ on the ring, emf induced in the element,

$de=\left(\frac{e}{2\pi a}\right)d\ell$

resistance of the element,
$dR=\left(\frac{R}{2\pi a}\right)d\ell$

$\therefore$ Potential difference across the element

$=de-idR$
$=\left(\frac{e}{2\pi a}\right)d\ell -\left(\frac{e}{R}\right)\left(\frac{R}{2\pi a}\right)d\ell =0$