Question

A solenoid 0.425 m long has 950 turns of wire. What is the magnetic field in the center of the solenoid when it carries a current of 2.75 A? (JC 19.57)

Answer 1

The magnetic field in the center of the solenoid is
`7.8*10^(-3)\ T`.

Step-by-step explanation:

Given that,

Length of solenoid = 0.425 m

Number of turns N = 950

Current I = 2.75 A

The magnetic field in the center of the solenoid is the product of the current , number of turns per unit length and permeability.

In mathematical form,

`B = \mu_(0)nI`

Where,
`n = (N)/(l)`

N = number of turns

L = length

I = current

Now, The magnetic field

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

Put the value into the formula

`B=(4\pi*10^(-7)*950*2.75)/(0.425)`

`B=(4*3.14*10^(-7)*950*2.75)/(0.425)`

`B=7.8*10^(-3)\ T`

Hence, The magnetic field in the center of the solenoid is
`7.8*10^(-3)\ T`.