Question

The current in a solenoid is 30A the number of turns per unit length is 500 turns per metre.Calculate the magnetic field if the core is air.

Answer 1

Given :

The magnitude of current in the solenoid = 30 A

The no of turns per unit length of the solenoid = 500 turns per metre

To Find :

The magnitude of magnetic field in the solenoid if the core is air = ?

Solution :

∴ We know that magnetic field of a solenoid is given as :

= $\mu_0 nI$ ( here $\mu_0$ is permeability of free space , n is no of turns per unit length and I is current )

So the magnetic field for the given solenoid is :

$B = \mu_0nI$

   =$4 \pi \times 10^{-7}\times 500 \times 30$

   =$188.4 \times 10^{-4} T$

   =$0.0188 T$

Hence the magnetic field of the solenoid is 0.0188 Tesla .

Answer 2

Answer:

B= 6π×10^-3

Explanation:

given:

B=uni

u=4π×10^-7

i=30A

n=500 turns

B=6π×10^-3