The inductance of a close-packed coil of 400 | vedclass

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Question

(A) The relationship between magnetic flux linkage and inductance is given by $N\phi = Li$,where $N$ is the number of turns,$\phi$ is the magnetic flu

Vedclass · Accepted Answer

$\frac{1}{4\pi} \mu_0 \, Wb$

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(A) The relationship between magnetic flux linkage and inductance is given by $N\phi = Li$,where $N$ is the number of turns,$\phi$ is the magnetic flux per turn,$L$ is the inductance,and $i$ is the current.Given: $N = 400$,$L = 8 \, mH = 8 	imes 10^{-3} \, H$,$i = 5 \, mA = 5 	imes 10^{-3} \, A$.Substituting the values into the formula:$\phi = \frac{Li}{N} = \frac{8 	imes 10^{-3} 	imes 5 	imes 10^{-3}}{400} = \frac{40 	imes 10^{-6}}{400} = 0.1 	imes 10^{-6} = 10^{-7} \, Wb$.We know that $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$. Therefore,$10^{-7} \, Wb = \frac{\mu_0}{4\pi} \, Wb$.

The inductance of a close-packed coil of $400$ turns is $8 \, mH$. $A$ current of $5 \, mA$ is passed through it. The magnetic flux through each turn of the coil is

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