A square loop of side 12 cm with its sides parallel to X and velocity of 8 cm s 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 T cm along the negative x-direction (that is it increases by 103 T cm- as one moves in the negative x-direction), and it is decreasing in time at the rate of 103 Ts. Determine the direction and magnitude of the induced current in axes is moved with a the loop if its resistance is 4.50 m2. Sol. Given, side of loop a = 12 cm . Area of loop (A) = a = (12) = 144 cm = 144 x 10 m? (. Area of square = (side))

Explanation

Given side of a loop a=12cm
`therefore` Area of loop `(A)=a^(2)=(12)^(2)=144cm^(2)=144xx10^(-4)m^(2) (therefore "Area of square"=("side")^(2))`
Velocity `v=8cm//s=8xx10^(2)m//s`
Rate of change of magnetic field with distance `(dB)/(dx)=10^(-3)T//cm`
Rate of change of magnetic field with time `(dB)/(dt)=10^(-3)T//s`
Resistance of the loop `R=4.5m Omega=4.5 xx 10^(-3)Omega`
Rate of change of magnetic flux with respect to time `(dphi)/(dt)=(d(BA))/(dt)=((dB)/(dt))A=10^(-3)xx144xx10^(-4)`
`=1.44xx10^(-5)Wb//s`
Rate of change of magnetic flux due to the motion of loop `(d theta)/(dt)=(dB)/(dx).A. (dx)/(dt)=10^(-3) xx 144xx10^(-4)xx8=11.52xx10^(-5)Wb//s`
Both of the effects cause a decrease in magnetic flux along the positive z-direction
Total induced emf in the loop `e=1.44xx10^(-5)+11/52xx10^(-5)`
`e=12.96 xx 10^(-5)V`
Induced current in the loop `=e/R=(12.96xx10^(-5))/(4.5xx10^(-3))=2.88xx10^(-2)A`
The direction of induced current is such as to increase the flux through the loop along positive z-direction.