Physics - Magnetism Question with Solution | TestHub

The magnetic field inside a long solenoid is given by $B={\mu }_{0}nI={\mu }_0\frac{N}{L}I$

The magnetic flux through the cross-section of the solenoid is $\Phi =BA={\mu }_0\frac{N}{L}IA$

The magnetic moment of the solenoid is $M=NIA$

Substituting $M$ into the flux equation, we get $\Phi =\frac{{\mu }_{0}M}{L}$

Rearranging for $M$, we get $M=\frac{\Phi L}{{\mu }_0}$

Substituting the given values $\Phi =1.57\times 10^{-6}\text{ Wb}$, $L=0.8\text{ m}$, and ${\mu }_0=4\pi \times 10^{-7}\text{ SI unit}$:

$M=\frac{1.57\times 10^{-6}\times 0.8}{4\times 3.14\times 10^{-7}}$

$M=\frac{1.256\times 10^{-6}}{12.56\times 10^{-7}}$

$M=1{\text{ Am}}^2$

Answer: $1{\mathrm{Am}}^2$