Question

A 0.50-m long solenoid consists of 500 turns of copper wire wound with a 4.0 cm radius. When the current in the solenoid is 22 A, the magnetic field at a point 1.0 cm from the central axis of the solenoid is

Answer 1

Answer: The magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.

Step-by-step explanation:

Given: Length = 0.50 m

No. of turns = 500

Current = 22 A

Formula used to calculate magnetic field is as follows.

`B = \mu_(o)((N)/(L))I`

where,

B = magnetic field

`\mu_(o)` = permeability constant =
`4\pi * 10^(-7) Tm/A`

N = no. of turns

L = length

I = current

Substitute the values into above formula as follows.

`B = \mu_(o)((N)/(L))I\\= 4 \pi * 10^(-7) Tm/A * ((500)/(0.5 m)) * 22\\= 0.0276 T`

Thus, we can conclude that magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.