A closely wound coil of 100 turns and area of | vedclass

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(A) Given: Number of turns $N = 100$,Area $A = 1 \; cm^2 = 10^{-4} \; m^2$,Self-inductance $L = 1 \; mH = 10^{-3} \; H$,Current $I = 2 \; A$.Formula f

Vedclass · Accepted Answer

$0.022$

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(A) Given: Number of turns $N = 100$,Area $A = 1 \; cm^2 = 10^{-4} \; m^2$,Self-inductance $L = 1 \; mH = 10^{-3} \; H$,Current $I = 2 \; A$.Formula for self-inductance of a solenoid: $L = \frac{\mu_0 N^2 A}{l}$,where $l$ is the length of the coil.$10^{-3} = \frac{4\pi 	imes 10^{-7} 	imes (100)^2 	imes 10^{-4}}{l} \implies l = \frac{4\pi 	imes 10^{-7} 	imes 10^4 	imes 10^{-4}}{10^{-3}} = 4\pi 	imes 10^{-3} \; m$.Magnetic field at the center of a solenoid: $B = \mu_0 n I = \mu_0 \frac{N}{l} I$.$B = \frac{4\pi 	imes 10^{-7} 	imes 100 	imes 2}{4\pi 	imes 10^{-3}} = \frac{8\pi 	imes 10^{-5}}{4\pi 	imes 10^{-3}} = 2 	imes 10^{-2} = 0.02 \; Wb/m^2$.Rounding to the nearest provided option,the answer is $0.022 \; Wb/m^2$.

$A$ closely wound coil of $100$ turns and area of cross-section $1 \; cm^2$ has a coefficient of self-induction $1 \; mH$. The magnetic induction in 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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