The figure shows a square loop L of side 5 c m which is connected to a network of resistances. The whole setup is moving towards the right with a constant speed of 1 c m s - 1 . At some instant, a part of L is in a uniform magnetic field of 1 T perpendicular to the plane of the loop. If the resistance of L is 1 . 7 Ω , the current in the loop at that instant will be close to:

Step-by-step solution

If B is magnetic field, v is velocity of rod and l is the length of rod, then induced EMF in moving road is given by ε = B v l = 1 T ( 1   c m   s - 1 ) ( 5   c m ) = 1 × 1 × 10 - 2 × 5 × 10 - 2 = 5 × 10 - 4    V As given circuit forms balanced whetstone bridge, the current through 3   Ω is zero. Equivalent resistance R e q = 4 × 2 4 + 2 + 1.7 = 3   Ω Current in the circuit i = ε R e q = 5 × 10 - 4 3 = 167   μ A