Question

A solenoid is 2.50 cm in diameter and 30.0 cm long. It has 300 turns and carries 12 A. (a) Calculate the magnetic field inside the solenoid. (b) Calculate the magnetic flux through the surface of a circle of radius 1.00 cm, which is positioned perpendicular to and centered on the axis of the solenoid. (c) The perimeter of this circle is made of conducting material. The current in the solenoid uniformly goes from 12 A to 10 A in 0.001 seconds. What e.M.F. Is generated in the conducting material? (d) Now, inside the solenoid we put a bar of steel, a ferromagnetic material. Steel has got a magnetic permeability μm = 4000μ0 . If the current in the solenoid is 12 A, what's the total magnetic field in the steel bar?

Answer 1

a) B = 0.0151 T

b)
`\theta = 4.737 * 10^-^6 Tm^2`

c)
`E = 7.896*10^-^4`

d)
`= 60.4 T`

Step-by-step explanation:

Given:

Diameter, d = 2.5 cm convert to meters = 0.025 m

Length, L = 30.0 cm convert to meters = 0.3 m

Number of turns, N = 300

Current, I = 12 A

a) Let's use the formula below to find the magnetic field.

`B = (u_0 N I)/(L)`

`B = (4\pi * 10^-^7 * 300 * 12)/(0.3)`

`B = (0.004524)/(0.3)`

B = 0.0151 T

b) Given a radius, r = 1.00 cm, let's convert it to meters = 0.01 m

Let's use the formula below to find magnetic flux.

`\theta = BA`

where
`A = \pi r^2`

Therefore,

`\theta = B \pi r^2`

`= 0.0151 \pi * (0.01)^2`

`\theta = 4.737 * 10^-^6 Tm^2`

c) Given that the current in the solenoid uniformly goes from 12 A to 10 A in 0.001 seconds, the EMF generated will be:

`E = (- change in \theta)/(change in t) = (-u_0 N)/(L) * \pi r^2 * (change in I)/(change in t)`

`E = (-4\pi * 10^-^7 * 300)/(0.3) * \pi *(0.01)^2 * (10 - 12)/(0.001)`

`-0.00125663 * \pi *(0.01)^2 * -2000`

`E = 7.896*10^-^4`

d) Let's use the formula:

`(u_m N I)/(L) = 4000 (u_0 NI)/(L)`

`4000 * 0.0151`

`= 60.4 T`