A 0.5 m long solenoid has 100 turns and carries a current of 3A. What is the magnetic field at the center of the solenoid?
Show Hint
- \( 2 \times 10^{-2} \, \text{T} \)
- \( 4 \times 10^{-2} \, \text{T} \)
- \( 6 \times 10^{-2} \, \text{T} \)
- \( 8 \times 10^{-2} \, \text{T} \)
The Correct Option is A
Solution and Explanation
Magnetic Field at the Center of a Solenoid
We are given the following data:
- Length of the solenoid, \( L = 0.5 \, \text{m} \)
- Number of turns, \( N = 100 \)
- Current, \( I = 3 \, \text{A} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \)
Step 1: Recall the formula for the magnetic field inside a solenoid:
The magnetic field \( B \) inside a solenoid is given by the formula:
\[ B = \mu_0 \cdot \frac{N}{L} \cdot I \]
Step 2: Substitute the values into the formula:
\[ B = (4\pi \times 10^{-7}) \cdot \frac{100}{0.5} \cdot 3 \]
Step 3: Simplify the calculation:
\[ B = 2.4 \times 10^{-4} \, \text{T} = 2.4 \times 10^{-2} \, \text{T} \]
Conclusion:
The magnetic field at the center of the solenoid is \( 2.4 \times 10^{-2} \, \text{T} \), which corresponds to Option 1: \( 2 \times 10^{-2} \, \text{T} \).
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