Question

A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1300 turns of wire. When running, the solenoid produced a field of 1.3 T in the center. Given this, how large a current does it carry?

Answer 1

1671 A

Step-by-step explanation:

With a logical assumption that magnetic permeability of vacuum would be used,

the magnetic field at the centre of a solenoid is given by

B = μ₀I (N/L)

Some textbooks use turn density, n = (N/L)

Either ways,

B = 1.3 T

μ₀ = (4π × 10⁻⁷) H/m

I = ?

N = 1300 turns

L = 2.1 m

I = (BL/μ₀N)

I = (1.3×2.1) ÷ (4π × 10⁻⁷ × 1300)

I = 1671 A

Hope this Helps!!!