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(D) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.Since magnetic flux $\phi = B \cdot A$, we have $B = \frac{\phi}{A}

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(D) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.Since magnetic flux $\phi = B \cdot A$, we have $B = \frac{\phi}{A}$.Equating the two expressions for $B$: $\frac{\phi}{A} = \frac{\mu_0 NI}{L}$.Rearranging to find the magnetic moment $M = NIA$: $NIA = \frac{\phi L}{\mu_0}$.Given $\phi = 1.57 	imes 10^{-6} \, Wb$, $L = 0.6 \, m$, and $\mu_0 = 4 \pi 	imes 10^{-7} \, T \cdot m/A$.Substituting the values: $M = \frac{1.57 	imes 10^{-6} 	imes 0.6}{4 \pi 	imes 10^{-7}}$.Using $\pi \approx 3.14$, $4 \pi \approx 12.56$.$M = \frac{1.57 	imes 10^{-6} 	imes 0.6}{12.56 	imes 10^{-7}} = \frac{0.942 	imes 10^{-6}}{12.56 	imes 10^{-7}} = \frac{9.42}{12.56} \approx 0.75 \, Am^2$.

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