Question

A solenoid of length 3.00 cm and radius 0.450 cm has 37 turns. If the wire of the solenoid has 1.95 amps of current, what is the magnitude of the magnetic field inside the solenoid

Answer 1

Answer: The magnitude of the magnetic field inside the solenoid is 3.02 x 10⁻³ T

Explanation:

l = 0.03 m

r = 0.00450 m

N = 37 turns

I = 1.95 A

μ = 4π x 10⁻⁷ TmA

We know that Magnetic field, B= μ x N x
`(I)/(L\\)`

Therefore B =
`(4\pi *10^(-7) * 1.95 * 37 )/(0.03)`

= 3.02 x 10⁻³ T