Question:
The magnetic field inside a 200 turns solenoid of radius 10 cm is \( 2.9 \times 10^{-4} \) Tesla. If the solenoid carries a current of 0.29 A, then the length of the solenoid is:
The magnetic field inside a 200 turns solenoid of radius 10 cm is \( 2.9 \times 10^{-4} \) Tesla. If the solenoid carries a current of 0.29 A, then the length of the solenoid is:
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For calculating the length of a solenoid, use the formula for the magnetic field inside a solenoid and solve for the length.
Updated On: Feb 2, 2026
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Correct Answer: 8
Solution and Explanation
Assuming a long solenoid, the magnetic field is given by the formula:
\[
B = \mu_0 \frac{N}{l} I
\]
Where:
- \( B = 2.9 \times 10^{-4} \, \text{T} \),
- \( N = 200 \) turns,
- \( I = 0.29 \, \text{A} \),
- \( l \) is the length of the solenoid,
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
Now, solving for \( l \): \[ l = \frac{\mu_0 N I}{B} = \frac{(4\pi \times 10^{-7})(200)(0.29)}{2.9 \times 10^{-4}} \, \text{m} \] \[ l = 8 \, \text{m} \] Thus, the length of the solenoid is 8 meters.
- \( B = 2.9 \times 10^{-4} \, \text{T} \),
- \( N = 200 \) turns,
- \( I = 0.29 \, \text{A} \),
- \( l \) is the length of the solenoid,
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
Now, solving for \( l \): \[ l = \frac{\mu_0 N I}{B} = \frac{(4\pi \times 10^{-7})(200)(0.29)}{2.9 \times 10^{-4}} \, \text{m} \] \[ l = 8 \, \text{m} \] Thus, the length of the solenoid is 8 meters.
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