The magnetic flux near the axis and inside the air core solenoid of length $60 \mathrm{~cm}$ carrying…
The magnetic flux near the axis and inside the air core solenoid of length $60 \mathrm{~cm}$ carrying current ' $I$ ' is $1.57 \times 10^{-6} \mathrm{~Wb}$. Its magnetic moment will be $\left[\mu_0=4 \pi \times 10^{-7}\right.$, SI unit and cross sectional area is very small as compared to length of solenoid.]
$1 \mathrm{Am}^2$
$0.25 \mathrm{Am}^2$
$0.5\mathrm{Am}^2$
$0.75 \mathrm{Am}^2$
Solution
Magnetic field inside the solenoid is given by
$\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\
& \therefore \frac{\phi}{\mathrm{A}}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\
& \therefore \text { Magnetic moment, NIA }=\frac{\phi \mathrm{L}}{\mu_0} \\
& =\frac{1.5 \times 10^{-6} \times 0.6}{4 \times 3.14 \times 10^{-7}} \\
& =0.75 \mathrm{Am}^2
\end{aligned}$