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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A metal rod of resistance $20 \;\Omega$ is fixed along the diameter of a conducting ring of radius$0.1 m$ . A magnetic field of $B=50T$ acts perpendicular to the plane The ring rotates with an angular velocity of $w=20\;rad/s$ about its axis . An external resistance of $10 \;\Omega$ is connected across center and rim of the ring. What is current through the external resistance.

$(a)\;1\;A\\ (b)\;\frac{1}{3}A \\(c)\;\frac{1}{2}\;A \\(d)\;\frac{1}{4}\;A $

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1 Answer

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Potential difference between centre and rim of the ring is
$V=\large\frac{1}{2}$$Bwr^2$
$\quad= \large\frac{1}{2} $$50. 20 \times (.1)^2$
$\quad =5V$
Now the equivalent circuit is :
$i=\large\frac{5}{10+5}$
$\quad= \large\frac{1}{3}$$A$
Hence b is the correct answer.
answered Mar 24, 2014 by meena.p
 

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