Question

A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid

Answer 1

1.37 ×10^-3 T

Step-by-step explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T