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

Answer:

Therefore,

Strength magnetic field at within the Solenoid,

$B =1.05\times 10^{-2}\ T$

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

$\mu_{0}=Permeability\ in\ free\ space=4\pi\times 10^{-7}\ Tm/A$

Therefore,

$B =\dfrac{\mu_{0}NI}{l}$

Substituting the values we get

$B =\dfrac{4\times 3.14\times 10^{-7}\times 1800\times 2.5}{0.54}= 1.05\times 10^{-2}\ T$

Therefore,

Strength magnetic field at within the Solenoid,

$B = 1.05\times 10^{-2}\ T$