A square metal loop of side 10 c m and resistance 1 Ω is moved with a constant velocity partly inside a uniform magnetic field of 2 W b m - 2 , directed into the paper, as shown in the figure. The loop is connected to a network of five resistors each of value 3 Ω . If a steady current of 1 m A flows in the loop, then the speed of the loop is -

Question

A square metal loop of side 10 c m and resistance 1 Ω is moved with a constant velocity partly inside a uniform magnetic field of 2 W b m - 2 , directed into the paper, as shown in the figure. The loop is connected to a network of five resistors each of value 3 Ω . If a steady current of 1 m A flows in the loop, then the speed of the loop is -, 0.5 cm s − 1, 1 cm s − 1, 2 cm s − 1, 4 cm s − 1

MARKS App · Accepted Answer

The network appears as a Balanced Wheatstone bridge. Therefore, no current will flow through the bridge resistance. Thus, the resistance will have no effect on the network. Set of two series resistors each values 3 Ω is now parallel to each other. Therefore the equivalent resistance of the balanced network is, ∴ R = 1 ( 1 3 + 3 + 1 3 + 3 ) = 1 ( 1 6 + 1 6 ) = 1 ( 2 6 ) = 6 2 = 3 Ω The balanced network is in series with the 1 Ω internal resistance. Therefore, the equivalant resistance of the network is, R ' = 3 + 1 = 4 Ω The direction of motion of free electron in rod in magnetic field is towards right and it has a velocity v. The induced emf is, e = v B l = I R ' Therefore, the velocity is, v = I R ' B l v = ( 1 × 10 − 3 ) ( 4 ) ( 2 ) ( 10 × 10 − 2 ) = 2 × 10 − 2 m s − 1 v = 2 cm s − 1

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