Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Induction

A conducting ring of radius R is placed in a magnetic field perpendicular to the plane of the coil at the rate of $\beta$. Then Which one is correct ?

$(a)\;\text{All points on ring will not be at same potential} \\(b)\;\text{emf induced the ring is $\large\frac{\pi}{2}$$ R^2 \beta$ }\\ (c)\;\text {electric intensity at any point is zero} \\ (d)\;E= \large\frac{R\beta}{2}$

1 Answer

$\phi= \pi R^2B$
$e= \pi R^2 \large\frac{dB}{dt}$
$\quad =\pi R^2 \beta$
current $=\large\frac{e}{r}$
Considering $dl$ a small element of ring.
$de=\bigg(\large\frac{e}{2 \pi R}\bigg)$$dl$
$de=\bigg(\large\frac{r}{2 \pi R}\bigg)$$dl$
potential difference $= de-idr$
$\qquad= \large\frac{e}{2 \pi R}$$ dl - \frac\large{e}{r} \frac{r}{2 \pi R}$$dR=0$
$\therefore$ all points in the ring are same potential
$\oint E.\bar{dl} =\large\frac{d \phi}{dt} $
$El= \pi R^2 \large\frac{dB}{dt}$
$E 2\pi R= \pi R^2 \beta$
$E=\large\frac{\beta R}{2}$
Hence d is the correct answer.
answered Mar 26, 2014 by meena.p

Related questions