Question

A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid cons

Answer 1

Complete question:

A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.

Answer:

The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

Explanation:

Given;

length of solenoid, L = 2.4 m

radius of solenoid, R = 1.7 cm = 0.017 m

current in the solenoid, I = 0.19 A

number of turns of the solenoid, N = 2100 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μnI

Where;

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

n is number of turns per length = N/L

I is current in the solenoid

B = μnI = μ(N/L)I

B = 4π x 10⁻⁷(2100 / 2.4)0.19

B = 4π x 10⁻⁷ (875) 0.19

B = 2.089 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.