Question

A long solenoid that has 1,140 turns uniformly distributed over a length of 0.415 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?

Answer 1

Therefore,

Current required is , I

`I = 0.0289\ Ampere`

Step-by-step explanation:

Given:

Turns = N = 1140

length of solenoid = l = 0.415 m

Magnetic Field,

`B = 1.00* 10^(-4)\ T`

To Find:

Current , I = ?

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,

`I =(Bl)/(\mu_(0)N)`

Substituting the values we get

`I =(1.00* 10^(-4)* 0.415)/(4* 3.14* 10^(-7)* 1140)=0.0289\ Ampere`

Therefore,

Current required is , I

`I = 0.0289\ Ampere`