Question

A square loop of side 12 cm with its sides parallel to X and Y axes ismoved with a velocity of 8 cm s–1 in the positive x-direction in anenvironment containing a magnetic field in the positive z-direction.The field is neither uniform in space nor constant in time. It has agradient of 10–3 T cm–1along the negative x-direction (that is it increasesby 10 – 3 T cm–1as one moves in the negative x-direction), and it isdecreasing in time at the rate of 10–3 T s–1. Determine the direction andmagnitude of the induced current in the loop if its resistance is 4.50 mW.

Answer 1

Given: The side of the square loop is 12 cm, the velocity of the loop is v, the gradient of the magnetic field along negative x-direction is 10 −3   Tcm -1 , the rate of decrease of the magnetic field is 10 −3   Ts -1 and the resistance of the loop is 4.50 mΩ.

The area of the loop is given as,

A= a 2

Where, the side of the square loop is a.

Since, 1 cm=0.01 m and 1 mΩ= 10 −3  Ω.

By substituting the given values in the above formula, we get

A=12× 10 −2 ×12× 10 −2 =144× 10 −4   m 2

The change in flux is given as,

dϕ dt =A× dB dx ×v

Where, the area of the square loop is A, the gradient of the magnetic field along negative x-direction and the velocity of the loop is v.

By substituting the given values in the above formula, we get

dϕ dt =144× 10 −4 × 10 −1 ×0.08 =11.52× 10 −5   Tm 2 s −1

The rate of change of the flux due to time variation is given as,

d ϕ ′ dt =A× dB dt

Where, the rate of decrease of magnetic field is dB dt .

By substituting the given values in the above expression, we get

d ϕ ′ dt =144× 10 −4 × 10 −3 =1.44× 10 −5   Tm 2 s −1

The induced emf in the loop is given as,

e= dϕ dt + d ϕ ′ dt

By substituting the given values in the above expression, we get

e=1.44× 10 −5 +11.52× 10 −5 =12.96× 10 −5  V

The induced current in the loop is given as,

i= e R

Where, the resistance of the loop is R.

By substituting the given values in the above formula, we get

i= 12.96× 10 −5 4.5× 10 −3 =2.88× 10 −2  A

Since, the flux is decreasing, then according to Lenz’s law the induced current should oppose the cause of the change. The induced current is induced in positive z direction.

Thus, the magnitude of the induced current is 2.88× 10 −2  Aand the direction of the current is positive z direction.