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:

Hint:

For calculating the length of a solenoid, use the formula for the magnetic field inside a solenoid and solve for the length.

Answer 1

Correct Answer: 8

For a long solenoid, the magnetic field \( B \) is expressed by the equation: \[ B = \mu_0 \frac{N}{l} I \] The parameters are:

• \( B = 2.9 \times 10^{-4} \, \text{T} \) (magnetic field strength)
• \( N = 200 \) (number of turns)
• \( I = 0.29 \, \text{A} \) (current)
• \( l \) (length of the solenoid)
• \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)

Rearranging the formula to solve for \( l \): \[ l = \frac{\mu_0 N I}{B} = \frac{(4\pi \times 10^{-7})(200)(0.29)}{2.9 \times 10^{-4}} \, \text{m} \] The calculated length of the solenoid is: \[ l = 8 \, \text{m} \] Therefore, the solenoid has a length of 8 meters.