A vertical ring of radius r and resistance R falls vertically. It is in contact with two vertical rails which are joined at the top, figure. The rails are without friction and resistance. There is a horizontal uniform magnetic field of magnitude B perpendicular to the plane of the ring and the rails. When the speed of the ring is $v$, the current is the section $PQ$ is [image:7_image28.emf]

Question

Dr. Abubacker · Accepted Answer

(d) When a ring moves in a magnetic field in a direction perpendicular to its plane, we replace the ring by a diameter (2r) perpendicular to the direction of motion. The emf is induced across this diameter. Current flow in the ring will be through the two semicircular portions in parallel.
Induced emf = $B\ (2\ r)v.$
Resistance of each half of ring = R /2
As the two halves are in parallel, therefore, equivalent resistance = R /4
$	herefore\ \ $ Current in the section $= \ \frac{B(2r)v}{R/4}$
$I = \frac{8Brv}{R}$

Explanation

Reviewed by Dr. Abubacker

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