The magnetic flux near the axis and inside the air core solenoid of...

The magnetic field inside a long solenoid is given by:

$B = \dfrac{\mu_0 N I}{L}$

where $N$ is the total number of turns, $I$ is the current, and $L$ is the length of the solenoid.

The magnetic flux through the cross-section of the solenoid is:

$\Phi = B A = \dfrac{\mu_0 N I A}{L}$

The magnetic moment of a solenoid is given by:

$M = N I A$

Substituting $M$ into the flux equation, we get:

$\Phi = \dfrac{\mu_0 M}{L}$

Rearranging to solve for the magnetic moment $M$:

$M = \dfrac{\Phi L}{\mu_0}$

Given values are:
$\Phi = 1.57 \times 10^{-6}\ \text{Wb}$
$L = 60\ \text{cm} = 0.6\ \text{m}$
$\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A} = 4 \times 3.14 \times 10^{-7}\ \text{T m/A} = 12.56 \times 10^{-7}\ \text{T m/A}$

Substituting these values into the formula:

$M = \dfrac{1.57 \times 10^{-6} \times 0.6}{12.56 \times 10^{-7}}$

$M = \dfrac{1.57}{12.56} \times 0.6 \times 10$

$M = \dfrac{1}{8} \times 6 = \dfrac{3}{4} = 0.75\ \text{A m}^2$

Answer: $0.75\ \text{Am}^2$