Question

A 0.5 m long solenoid has 500 turns and has a flux density of \(2.52 \times 10^{-3}\) T at its center. The current in the solenoid is (Given, \(\mu_0 = 4\pi \times 10^{-7}\) H/m)

• A. 1.2 A
• B. 2.0 A
• C. 2.8 A
• D. 3.4 A

Hint:

Use \(B=\mu_0 n I\) with \(n=N/L=1000\) per m.

Answer 1

The Correct Option is B

Step 1: Set up.
Let the current be \(I\). For a long solenoid the axial field is \(B=\mu_0 (N/L) I\).

Step 2: Rearrange first.
Making \(I\) the subject gives \(I = \dfrac{BL}{\mu_0 N}\).

Step 3: Put in numbers.
\[ I = \frac{2.52\times10^{-3}\times 0.5}{4\pi\times10^{-7}\times 500} = \frac{1.26\times10^{-3}}{6.283\times10^{-4}} \]

Step 4: Finish.
This gives \(I \approx 2.005\) A, which rounds to 2.0 A. That is option 2.

Final Answer:
The solenoid carries 2.0 A, option 2. \[ \boxed{2.0 \text{ A}} \]