Question

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside the solenoid near its center?

Answer 1

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Step-by-step explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

`B = (\mu_o NI)/(l) \\\\`

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

`B = (4\pi*10^(-7) *269*3.9)/(1.02) \\\\B = 1.293 *10^(-3) \ T`

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T