Question

A current of 6.67 A in a solenoid of length 13.6 cm creates a 0.373 T magnetic field at the center of the solenoid. How many turns does this solenoid contain?

Answer 1

Answer:

The value is  $N = 6051 \ turns$

Explanation:

From the question we are told that

   The current is  $I = 6.67 \ A$

  The length is  $l = 13.6 \ cm = 0.136 \ m$

  The magnetic field is  $B = 0.373 \ T$

Gnerally the number of turn per unit length is mathematically represented as

      $n = \frac{B}{\mu_o * I}$

Here $\mu_o$ is the permeability of free space with value  $\mu_o = 4*\pi*10^{-7}$

So

=>    $n = \frac{0.373}{ 4\pi * 10^{-7} * 6.67 }$

=>    $n = 44496 \ turns /m$

Generally the number of turn is mathematically represented as

       $N = n * l$

=>  $N = 44496 * 0.136$

=>  $N = 6051 \ turns$

Answer 2

The no of turns dos this solenoid contains will be 6051 turns.

What is a solenoid?

A coil of wire that carries an electric current is a solenoid. A solenoid is an electromagnet formed by a helical coil of wire. Which generates a magnetic field when an electric current is passed through the coil .

The given data in the problem is;

B is the current =6.67 A

L is the length =13.6 cm=0.136 m

B is the magnetic field =0.373 T  

The number of turns per unit length is found as;

$\rm n= \frac{B}{\mu_0I} \\\\ \rm n= \frac{0.373}{4 \times 3.14 \times 10^{-7}\times 6.67} \\\\ \rm n = 44496 \ turn /m$

The number of turns is found as;

$\rm N=n\times l \\\\ \rm N=44496\times 0.136 \\\\ \rm N=6051 \ turns$

Hence the no of turns dos this solenoid contains will be 6051 turns.

To learn more about the solenoid refer to the link ;

brainly.com/question/16015159