A copper wire of length 40 cm, diameter 2mm and resistivity `1.7xx10^(-8)Omega m` form a square frame. If a uniform magnetic field B exists in a direction perpendicular to the plane of square frame and it changes at a steadyrate `(dB)/(dt)=0.02` T/s, then find the current induced in the frame.

Options

A. `9.3xx10^(-2)` amp(b)
B. `9.3xx10^(-1)` amp
C. `3.3xx10^(-2)` amp(d)
D. `19.3xx10^(-2)` amp

Answer

A

Explanation

Area of the loop `=0.1xx0.1=0.01 m^(2)`
`epsilon=-(d phi)/(dt)=(-d)/(dt)(BA)`
Magnitude of emf
`epsilon = A(dB)/(dt)=(0.01m^(2))(0.02 T//s)=2xx10^(-4)V`
Resistance of the loop is
`R=rho(l)/(A)=(1.7xx10^(-8)xx40xx10^(-2))/(3.14xx10^(-6))=2.16xx10^(-3)Omega`
Current induced in the loop
`I=(epsilon)/(R )=(2xx10^(-4)V)/(2.16xx10^(-3)Omega)=9.3xx10^(-2)` amp.