Question

The magnetic flux near the axis and inside the air core solenoid of length 60 cm carrying current I is \(\frac{π}{2}\times 10^{-6}\) Wb. Its magnetic moment will be (cross-sectional area is very small as compared to length of solenoid, \(μ_0\) = Permeability of free space = \(4π\times 10^{-7}\) SI unit)

• A. \(0.75\text{ Am}^2\)
• B. \(0.25\text{ Am}^2\)
• C. \(0.50\text{ Am}^2\)
• D. \(1.0\text{ Am}^2\)

Hint:

Use B = mu naught n I and the magnetic moment m = N I A, then eliminate n, I and A using the flux.

Answer 1

The Correct Option is A

Step 1: Think of the moment as a bar magnet.
For a solenoid, $m = \dfrac{B}{\mu_0}\times\text{Volume}$, since $B = \mu_0 nI$ and $m = nIAL$.

Step 2: Volume in terms of flux.
$B\times A = \Phi$, so $m = \dfrac{\Phi L}{\mu_0}$.

Step 3: Compute.
$m = \dfrac{(\pi/2\times10^{-6})(0.6)}{4\pi\times10^{-7}} = \dfrac{0.3}{0.4} = 0.75\text{ A m}^2$.

Final Answer:
Option (A). \[ \boxed{0.75\text{ A m}^2} \]