A circular coil of radius 5, cm has 500 turns | vedclass

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(A) The magnetic flux linked with a coil is given by $\phi = Li$,where $L$ is the self-inductance and $i$ is the current.Also,$\phi = NBA$,where $N$ i

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$25\, mH$

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(A) The magnetic flux linked with a coil is given by $\phi = Li$,where $L$ is the self-inductance and $i$ is the current.Also,$\phi = NBA$,where $N$ is the number of turns,$B$ is the magnetic field,and $A$ is the area of the coil.For a circular coil,the magnetic field at the center is $B = \frac{\mu_0 Ni}{2r}$.The area of the coil is $A = \pi r^2$.Equating the two expressions for flux: $Li = N \left( \frac{\mu_0 Ni}{2r} ight) (\pi r^2)$.Simplifying,we get $L = \frac{\mu_0 N^2 \pi r}{2}$.Given: $N = 500$,$r = 0.05\, m$,and $\mu_0 = 4\pi 	imes 10^{-7}\, T\cdot m/A$.Substituting the values: $L = \frac{4\pi 	imes 10^{-7} 	imes (500)^2 	imes \pi 	imes 0.05}{2}$.$L = 2 	imes 10^{-7} 	imes 250000 	imes \pi^2 	imes 0.05$.Using $\pi^2 \approx 10$,$L \approx 2 	imes 10^{-7} 	imes 250000 	imes 10 	imes 0.05 = 0.025\, H = 25\, mH$.

$A$ circular coil of radius $5\, cm$ has $500$ turns of a wire. The approximate value of the coefficient of self-induction of the coil will be:

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