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(A) Given: $N = 100$, $A = 1 \,cm^2 = 1 	imes 10^{-4} \,m^2$, $L = 1 \,mH = 1 	imes 10^{-3} \,H$, and $I = 2 \,A$. Self-inductance of a solenoid is

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(A) Given: $N = 100$, $A = 1 \,cm^2 = 1 	imes 10^{-4} \,m^2$, $L = 1 \,mH = 1 	imes 10^{-3} \,H$, and $I = 2 \,A$. Self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$, where $l$ is the length of the coil. From this, the length $l$ is: $l = \frac{\mu_0 N^2 A}{L} = \frac{(4 \pi 	imes 10^{-7}) 	imes (100)^2 	imes (1 	imes 10^{-4})}{1 	imes 10^{-3}} = 4 \pi 	imes 10^{-3} \,m$. The magnetic induction $B$ at the centre of the solenoid is given by $B = \frac{\mu_0 N I}{l}$. Substituting the value of $l$: $B = \frac{(4 \pi 	imes 10^{-7}) 	imes 100 	imes 2}{4 \pi 	imes 10^{-3}} = 10^{-7} 	imes 200 	imes 10^3 = 0.2 \,Wb/m^2$.

$A$ closely wound coil of $100$ turns and of cross-section $1 \,cm^2$ has a coefficient of self-inductance $1 \,mH$. The magnetic induction at the centre of the core of the coil when a current of $2 \,A$ flows in it will be (in $Wb/m^2$):

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