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 `mOmega`.

Explanation

Given, length of side of square loop, l = 12 cm = 0.12 m
Area of square loop, `A=l^(2) = 0.12 xx 0.12 = 144 xx 10^(-4) m^(2)`
Velocity of square loop, `v = (dx)/(dt) = 8 cms^(-1) = 0.08 ms^(-1)`
Gradient of magnetic field along the negative X-direction is
`(dB)/(dx) = -10^(-3) Tcm^(-1)= -10^(-1) T m^(-1)`
Rate of change of magnetic field with time is
`(dB)/(dt) = -10^(-3) T s^(-1)`
Rate of change of flux due to time variation in B
`((dphi)/(dt))_(1)=(d)/(dt)(BA)=A(dB)/(dt)`
`= 144 xx 10^(-4) xx 10^(-3)`
`= 144 xx 10^(-5) Wb s^(-1)`
Rate of change of flux due to motion of the loop in a non-uniform magnetic field (B)
`((dphi)/(dt))_(2)=A((dB)/(dx) xx (dx)/(dt))=A(dB)/(dx)v`
`= 144 xx 10^(-4) xx 10^(-1) xx 0.08`
`= 11.52 xx 10^(-5) Wb s^(-1)`
The two effects add up as both of them cause a decrease in flux along the positive Z-direction. Therefore, induced emf will be equal to total rate of change of flux linked with the square loop,
`epsilon=(1.44 xx 10^(-5)+11.52 xx 10^(-5))`
`epsilon=12.96 xx 10^(-5) V`
Induced current in the loop is
`I=(epsilon)/(R)=(12.96 xx 10^(-5))/(4.5 xx 10^(-3))=2.88 xx 10^(-2)`
As the loop moves in the field in positive X-direction along which field is decreasing, flux through coil decreases. Thus induced current will flow so as to increase the flux. The magnetic field due to induced current will support the applied magnetic field. It will be along positive Z-direction. The field will point out of the face of loop (towards observer). Therefore, the induced current will be anticlockwise.