A metallic square of loop ABCD of size 15 cm and Resistance 1 ohm is moved at uniform velocity of vm/s in a uniform magnetic field of 2TesIa the field lines being normal to the plane of paper. The loop is connected to an electrical network of customs, each of resistance 2 ohms. Calculate the Speed of the loop, for which 2mA Current fun in loop.
Asked by Rohan Astik
| 5th Jun, 2014,
07:43: AM
Expert Answer:
The network of resistors shown in the figure is a balanced wheatstone's bridge. Hence, its equivalent resistance R1 is given by,
As resistance of loop R2=1
, hence total resistance R = R1 + R2 = 1 + 2 = 3.
As length of side AB of loop l =15 cm = 0.15 m and B = 2 T and direction of motion v is perpendicular to B as well as l, hence induced emf
, hence total resistance R = R1 + R2 = 1 + 2 = 3.
Answered by Faiza Lambe
| 5th Jun, 2014,
10:53: AM
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