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(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$, where $n = N/L$ is the number of turns per unit length.
The magnetic flux $

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$, where $n = N/L$ is the number of turns per unit length.
The magnetic flux $\phi$ through a cross-sectional area $A$ is $\phi = B \cdot A = \mu_0 \frac{N}{L} I A$.
The magnetic moment $M$ of a solenoid is given by $M = N I A$.
Substituting $M$ into the flux equation: $\phi = \frac{\mu_0 M}{L}$.
Given $\phi = \frac{\pi}{2} 	imes 10^{-6} 	ext{ Wb}$, $L = 0.6 	ext{ m}$, and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.
$\frac{\pi}{2} 	imes 10^{-6} = \frac{4\pi 	imes 10^{-7} 	imes M}{0.6}$.
$\frac{1}{2} 	imes 10^{-6} = \frac{4 	imes 10^{-7} 	imes M}{0.6}$.
$0.5 	imes 10^{-6} = \frac{4 	imes 10^{-7} 	imes M}{0.6}$.
$M = \frac{0.5 	imes 10^{-6} 	imes 0.6}{4 	imes 10^{-7}} = \frac{0.3 	imes 10^{-6}}{4 	imes 10^{-7}} = \frac{0.3}{0.4} = 0.75 	ext{ Am}^2$.

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