Question

A long solenoid has a length of 0.54 m and contains 1900 turns of wire. There is a current of 2.5 A in the wire. What is the magnitude of the magnetic field within the solenoid?

Answer 1

Therefore,

Strength magnetic field at within the Solenoid,

`B =1.05* 10^(-2)\ T`

Step-by-step explanation:

Given:

Turn = N = 1900

length of solenoid = l = 0.54 m

Current, I = 2.5 A

To Find:

Strength magnetic field within the Solenoid,

B = ?

Solution:

If N is the number of turns in the length, the total current through the rectangle is NI. Therefore, Ampere’s law applied to this path gives

`\int {B} \, ds= Bl=\mu_(0)NI`

Where,

B = Strength of magnetic field

l = Length of solenoid

N = Number of turns

I = Current

`\mu_(0)=Permeability\ in\ free\ space=4\pi* 10^(-7)\ Tm/A`

Therefore,

`B =(\mu_(0)NI)/(l)`

Substituting the values we get

`B =(4* 3.14* 10^(-7)* 1800* 2.5)/(0.54)= 1.05* 10^(-2)\ T`

Therefore,

Strength magnetic field at within the Solenoid,

`B = 1.05* 10^(-2)\ T`