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: Understanding the Concept:
Inside a long solenoid, \(B = \mu_0 nI\), where \(n\) is the number of turns per unit length. The magnetic moment is \(m = NIA\), where \(N = nL\).

Step 2: Link flux and moment.
\(\Phi = BA = \mu_0 nIA\). Then
\[ m = NIA = nL\,IA = \frac{L}{\mu_0}\,(\mu_0 nIA) = \frac{L\Phi}{\mu_0} \]

Step 3: Substitute.
\(L = 0.6\) m, \(\Phi = \dfrac{\pi}{2}\times 10^{-6}\) Wb, \(\mu_0 = 4\pi\times 10^{-7}\).
\[ m = \frac{0.6\times\frac{\pi}{2}\times 10^{-6}}{4\pi\times 10^{-7}} = \frac{0.3\pi\times 10^{-6}}{4\pi\times 10^{-7}} = \frac{3}{4} = 0.75\text{ A m}^2 \]

Step 4: Check the options.
Only option (A), 0.75 A m\(^2\), matches.

Final Answer:
The magnetic moment is \(0.75\text{ A m}^2\). \[ \boxed{0.75\text{ A m}^2} \]