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 move 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.
Solution
Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12 × 0.12 = 0.0144 m2
Velocity of the loop, v = 8 cm/s = 0.08 m/s
Gradient of the magnetic field along negative x-direction,
`("dB")/("dx")` = 10−3 T cm−1 = 10−1 T m−1
And, rate of decrease of the magnetic field,
`("dB")/("dt")` = 10−3 T s−1
Resistance of the loop, R = 4.5 mΩ = 4.5 × 10−3 Ω
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
`("d"phi)/("dt") = "A" xx ("dB")/("dx") xx "v"`
= 144 × 10−4 m2 × 10−1 × 0.08
= 11.52 × 10−5 T m2 s−1
Rate of change of the flux due to explicit time variation in field B is given as:
`("d"phi"'")/("dt") = "A" xx ("dB")/("dt")`
= 144 × 10−4 × 10−3
= 1.44 × 10−5 T m2 s−1
Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:
e = 1.44 × 10−5 + 11.52 × 10−5
= 12.96 × 10−5 V
∴ Induced current, i = `"e"/"R"`
= `(12.96 xx 10^-5)/(4.5 xx 10^-3)`
i = 2.88 × 10−2 A
Hence, the direction of the induced current is such that there is an increase in the flux through the loop along the positive z-direction.
