The magnetic flux near the axis and inside th | vedclass

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(C) The magnetic induction $B$ inside a long solenoid is given by $B = \frac{\mu_0 N I}{L}$.Magnetic flux $\phi$ through the cross-sectional area $A$

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(C) The magnetic induction $B$ inside a long solenoid is given by $B = \frac{\mu_0 N I}{L}$.Magnetic flux $\phi$ through the cross-sectional area $A$ is $\phi = B A = \frac{\mu_0 N I A}{L}$.The magnetic moment $M$ of the solenoid is defined as $M = N I A$.Substituting $M$ into the flux equation,we get $\phi = \frac{\mu_0 M}{L}$.Rearranging for $M$,we have $M = \frac{\phi L}{\mu_0}$.Given $\phi = 1.57 	imes 10^{-6} \ Wb$,$L = 0.6 \ m$,and $\mu_0 = 4\pi 	imes 10^{-7} \ T \ m/A$.$M = \frac{1.57 	imes 10^{-6} 	imes 0.6}{4 	imes 3.14 	imes 10^{-7}}$.$M = \frac{1.57 	imes 10^{-6} 	imes 0.6}{12.56 	imes 10^{-7}} = \frac{0.942 	imes 10^{-6}}{1.256 	imes 10^{-6}} = 0.75 \ A \ m^2$.

The magnetic flux near the axis and inside the air-core solenoid of length $60 \ cm$ carrying current $I$ is $1.57 \times 10^{-6} \ Wb$. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length,$\mu_0 = 4\pi \times 10^{-7} \ SI \ units$). (in $A \ m^2$)

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