• # question_answer 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. 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 in the section PQ is A)  0                                B)  $\frac{2Brv}{R}$C)  $\frac{4Brv}{R}$                 D)  $\frac{8Brv}{R}$

Solution :

When a ring moves in a magnetic field perpendicular to its plane, replace the ring by a diameter perpendicular to the direction of motion. The emf is induced across this diameter. Induced emf $e=B(2r)v$ Resistance of each half of ring$=R/2$ As these are in parallel, the equivalent resistance = R/4 Current in the circuit$=\frac{B(2r)v}{R/4}=\frac{8Brv}{R}$ Hence, the correction option is [d].

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