Question

A solenoid with a certain number of turns N and carrying a current of 2.000 A has a length of 34.00 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mT, what is the value of N?

Answer 1

The number of turns of the solenoid is 1217 turns

Step-by-step explanation:

Given;

current in the solenoid, I = 2 A

length of the solenoid, L = 34 cm = 0.34 m

magnitude of the magnetic field, B = 9 mT = 0.009 T

Number of turns of the solenoid = N

The magnitude of magnetic field at the center of the solenoid is given by;

B = μnI

Where;

μ is permeability of free space = 4π x 10⁻⁷ m/A

I is the current in the solenoid

n is the number of turns per length

n = B/μI

n = (0.009) / (4π x 10⁻⁷)(2)

n = 3580.52 turns/m

N = nL

N =(3580.52 turns/m) x (0.34 m)

N = 1217 turns

Therefore, the number of turns of the solenoid is 1217 turns