6.12 A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10–3 T cm–1 along the negative x-direction (that is it increases by 10 – 3 T cm–1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10–3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

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8 months ago

6.12 Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 *0.12=0.0144m2

Velocity of the loop, v = 8 cm /s = 0.08 m/s

Gradient of the magnetic field along negative x – direction.

dBdx = 10-3 T/cm = 10-1 T/m

Rate of decrease of magnetic field

dBdt = 10-3 T/s

Resistance of the loop, R = 4.50 mΩ = 4.5 *10-3 Ω

Rate of change

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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