A solenoid with 400 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the magnitude of the magnetic field at the center of the solenoid?

The magnitude of the magnetic field inside the solenoid is 1.5 × 10⁻²T

Explanation:

Given that,

Length = 40 cm

Radius = 0.040 cm

Number of turns = 400

Current = 12 A

We need to calculate the magnetic field

Using formula of magnetic field inside the solenoid

Where,=Number of turns per unit length

I = current

B = magnetic field

Put the value into the formula

Hence, The magnitude of the magnetic field inside the solenoid is 1.5 × 10⁻²T

friction

explanation:

they us to move forward and backwards