Question

A solenoid 81.0 cm long has a radius of 1.70 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answer 1

The magnitude of the magnetic field inside the solenoid is
`7.3*10^(-3)\ T`.

Step-by-step explanation:

Given that,

Length = 81.0 cm

Radius = 1.70 cm

Number of turns = 1300

Current = 3.60 A

We need to calculate the magnetic field

Using formula of magnetic field inside the solenoid

`B =\mu nI`

`B =\mu(N)/(l)I`

Where,
`(N)/(l)`=Number of turns per unit length

I = current

B = magnetic field

Put the value into the formula

`B =4\pi*10^(-7)*(1300)/(81.0*10^(-2))*3.60`

`B = 7.3*10^(-3)\ T`

Hence, The magnitude of the magnetic field inside the solenoid is
`7.3*10^(-3)\ T`.