Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
0 votes

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}$

Can you answer this question?

1 Answer

0 votes
$\phi= \pi R^2B$
$e= \pi R^2 \large\frac{dB}{dt}$
$\quad =\pi R^2 \beta$
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

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App