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: Understanding the concept.
The field inside a long solenoid is uniform. It depends on the number of turns per unit length and on the current.

Step 2: Key formula.
\[ B = \mu_0 n I \] where \(n = N/L\) is the number of turns per metre.

Step 3: Find n.
\[ n = \frac{500}{0.5} = 1000 \text{ turns/m} \]

Step 4: Solve for I.
\[ I = \frac{B}{\mu_0 n} = \frac{2.52 \times 10^{-3}}{4\pi \times 10^{-7} \times 1000} = \frac{2.52 \times 10^{-3}}{1.2566 \times 10^{-3}} \approx 2.0 \text{ A} \]

Step 5: Check the options.
1.2 A, 2.8 A and 3.4 A do not match the value 2.005 A. Only 2.0 A matches, which is option 2.

Final Answer:
The current in the solenoid is 2.0 A. \[ \boxed{2.0 \text{ A}} \]