A uniform magnetic field B exists in a direction perpendicular to the plane of a square frame made of copper wire. The wire has a diameter of 2 mm and a total length of 40cm. The magntic field changes with time at a steady rate `dB/dt = 0.02 T s ^(-1)`. Find the current induced in the frame. Resistivity of copper `=1.7 X 10 ^(-8) Wm`.

Explanation

The area A of the loop `= ((40cm)/(4))((40cm)/(4)) =0.01m^2`
if the magnetic field at an instant is B, the flux through the frame at that instant will be `Omega = BA`
As the area remains constant, the magnitude of the emf induced will be ` epsilon=(dPhi)/(dt) = A(dB)/(dt)`
The resistance of the loop is `R =rho(i)/(A)`
`=((1.7X10^(-8) Omegam) (40cm)/(piX1 mm^2)`
`((1.7X10^(-8) Omega m)(40X10^(-2)m))/(3
14X1X10^(-6)m^2) = 2
16X 10(-6) m^2) =2
16 X 10^(-3) Omega`
Hence, the current induced in the loop will be `i =(2X10(-4)V)/(2
16X10^(-3) Omega) = 9
3 X 10^(-2) A