logo

Ask Questions, Get Answers

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

A conducting circular loop is placed in a uniform field of $B=0.05\;T$ with its plane perpendicular to the loop. The radius of loop is made to shrink at constant rate of $0.5\;mm$ per second. The induced emf when radius is $2 \;cm$ is

$(a)\;1 \times 10^{-6} \pi V \\ (b)\;2 \times 10^{-3} \pi V \\(c)\;1 \times 10^{-6} V \\(d)\;2 \times 10^{-3} V$

Can you answer this question?
 
 

1 Answer

0 votes
Induced emf $e= \large\frac{d\phi}{dt}$
Where $\phi$ is flux linked with the coil which is BA.
$\qquad= B \pi r^2$
$B= 0.05 \;T$
$\large\frac{dr}{dt} $$=0.5 \times 10^{-3}\;ms^{-1}$
$r= 2 \times 10^{-2}\;m$
$e= \large\frac{d \phi}{dt} =\frac{d}{dt} $$B \pi r^2$
$\qquad=B \pi 2r \large\frac{dr}{dt}$
$\qquad= 0.5 \pi \times 2 \times 2 \times 10^{-2} \times 0.5 \times 10^{-3}$
$\qquad= \pi \times 10^{-3} \times 10^{-2} \times 10^{-1}$
$\qquad= \pi \times 10^{-6}\;V$
Hence a is the correct answer.

 

answered Mar 21, 2014 by meena.p
edited Sep 24, 2014 by thagee.vedartham
 

Related questions

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