Question

A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?

Answer 1

The current is
`I = 2042\ A`

Step-by-step explanation:

From the question we are told that

The length of the solenoid is
`l = 2.2 \ m`

The radius is
`r_i = 30 \ cm = 0.30 \ m`

The number of turn is
`N = 1200 \ turns`

The magnetic field is
`B = 1.4 \ T`

The magnetic field produced is mathematically represented as

`B = (\mu_o * N * I )/(l )`

making
`I` the subject

`I = (B * l)/(\mu_o * N )`

Where
`\mu_o` is the permeability of free space with values
`\mu_o = 4\pi *10^(-7) N/A^2`

substituting values

`I = (1.4 * 2.2 )/(4\pi *10^(-7) * 1200 )`

`I = 2042\ A`