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(A) Given: Number of turns $N = 70$,radius $r = 10 \,cm = 0.1 \,m$,magnetic field $B = 2 	imes 10^{-3} \,T$,current $I = 2.2 \,A$.Since the magnetic

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(A) Given: Number of turns $N = 70$,radius $r = 10 \,cm = 0.1 \,m$,magnetic field $B = 2 	imes 10^{-3} \,T$,current $I = 2.2 \,A$.Since the magnetic field is perpendicular to the plane of the coil,the angle between the area vector and the magnetic field is $	heta = 0^{\circ}$.The magnetic flux linked with the coil is $\phi = N B A \cos 	heta$.Substituting the values: $\phi = 70 	imes (2 	imes 10^{-3}) 	imes (\pi 	imes (0.1)^2) 	imes \cos 0^{\circ}$.$\phi = 140 	imes 10^{-3} 	imes \pi 	imes 0.01 = 1.4 \pi 	imes 10^{-3} \,Wb$.Using $\pi \approx 3.14$,$\phi = 1.4 	imes 3.14 	imes 10^{-3} \approx 4.4 	imes 10^{-3} \,Wb$.For the net flux to vanish,the flux due to the current in the coil must equal the external flux: $\phi = L I$.$L = \frac{\phi}{I} = \frac{4.4 	imes 10^{-3}}{2.2} = 2 	imes 10^{-3} \,H = 2 \,mH$.

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