Question

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?

Answer 1

Answer:

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

$B =\mu nIB =\mu\dfrac{N}{l}I$

Where,$\dfrac{N}{l}$=Number of turns per unit length

I = current

B = magnetic field

Put the value into the formula

$B =4\pi\times10^{-7}\times\dfrac{400}{40\times10^{-2}}\times12B = 1.5\times10^{-2}\ T$

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