Question

A 44.5 mA current is carried by a uniformly wound air-core solenoid with 500 turns, a 18.5 mm diameter, and 14.0 cm length. (a) Compute the magnetic field inside the solenoid. µT (b) Compute the magnetic flux through each turn. T·m2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.)

Answer 1

a) 199.716 μT

b)
`5.368 * 10^(-8)` T·m^2

c) 0.603 mH

d) B and Ф

Step-by-step explanation:

I am giving the explanation with my handwritten solution in the paper.

Check the attachment please.

Answer 2

a. Magnetic Field =1.997×
`10^(-4)` T

b. Area= 2.68×
`10^(-4)`
`m^(2)`

Magnetic Flux= 5.367×
`10^(-8)`T
`m^(2)`

c. Inductance= 6.013×
`10^(-4)`H

Step-by-step explanation:

Parameters from the question

I= 44.5×
`10^(-3)`A

N=500 turns

Diameter=18.5mm

Radius = (diameter/2) = 9.25mm =9.25×
`10^(-3)`m

L= 14cm = 0.14m

Permitivity
`U_(o)`=4π×
`10^(-7)`H/m

The Formulars Used are

B(Magnetic Field) =
`(U_(o). N. I )/(l)`

Mag Flux= B.A

Inductance=
`(U_(o).N^(2) .A )/(l)`