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Q6.14 (d) Figure 6.20 shows a metal rod $PQ$ resting on the smooth rails $AB$ and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer $G$ connects the rails through a switch $K$ . Length of the $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown. Give the polarity and magnitude of the induced emf.What is the retarding force on the rod when K is closed?

Answers (1)

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The formula to calculate the force on the conductor moving in a magnetic field is,

$F=Bil$

Where,

$i$ is the induced current.

$B$ is the magnetic field strength.

$l$ is the length of the conductor.

The induced current is calculated as,

$I=e/R =\frac{9\times10^{-3}}{9\times10^{-3}}= 1A$

The length of the rod is 0.15 m.

Magnetic field strength, B=0.50 T.

Substitute the value in the above formula,

$F=Bil$ $=1\times 0.5\times 0.15 =75\times 10^{-3} N$

Hence, the retarding force on the rod is 75× 10 −3 N.

Posted by

Pankaj Sanodiya

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