Question

The magnetic intensity H at the centre of a long solenoid carrying a current of 2.0 A, is found to be 1500 A m−1. Find the number of turns per centimetre of the solenoid.

Answer 1

Explanation:

Step 1:

Given data  

Solenoid Current, I = 2 A

At center of long solenoid, magnetic intensity H = 1500 Am−1

A solenoid created magnetic field

Step 2:

B is provided

$B=\mu_{0} n i \quad \ldots e q^{n}(1)$

Where, n = Number of turns in length per segment

i = Electric current via a solenoid

Also, the relation between intensity of the magnetic field B and intensity  of magnetic

Step 3:

H is known as  

$H=\frac{B}{\mu_{0}} \quad \ldots e q^{n}(2)$

Step 4

We get from equations (1) and (2):

$H=n i$

$1500 \mathrm{A} / \mathrm{m}=n \times 2$

n = 750 turns/meter

n = 7.5 turns/cm

Answer 2

Answer:

n=7.5 turns/cm is ur answer ....