Question

1. A long skinny solenoid with a radius of 2 cm and a length of 80 cm is made with 10,000 turns of wire.

a. Calculate the magnitude of the field in the solenoid when it carries a current of 37.9 mA.

b. The solenoid is now filled with iron that has an average relative permeability of 500. Calculate the total magnetic field in the solenoid.

c. Determine the magnetic susceptibility and the magnetization in the iron.

Answers: a.) 595 micro tesla, b.) 298 micro tesla, c.) 499, 2.36*10^5 A/m

Answer 1

given

radius of the solenoid r = 2cm = 0.02m

length of the solenoid l = 80cm = 0.8m

no of turns N = 10,000 turns

1)    The formula is given by

B= $\mu$0nI ------------(1)

WHERE n is the number of turns per unit length given by n = N/l

n = 10000/0.8 = 12500 turns

here $\mu$0 = 4$\pi$ x 10^-7 H/m [ henr/metre]

current I = 37.9mA = 0.0379A

substituting in eqn (1) we get the value of magnetic firld in the solenoid as

B = 5.95 x 10^-4 T (tesla which is the unit of magnetic field)

2) Relative permeability is given by $\mu$r and it is defined as the ratio of magnetic permeability of the material medium ($\mu$) to the magnetic permeability of vaccum ($\mu$r).The formula is

    $\mu$r = $\mu$ / $\mu$0-------------(2)

here $\mu$r = 500 (given)

substituting in eqn (2) we get

500(4x3.14x10^-7) = $\mu$

so $\mu$ = 6.28 x 10^-4

substituting the above value in the formula

B = $\mu$ n I

= (6.28 x 10^-4) ( 12500)(0.0379)

   B = 297.51 x 10^-3 T

3) Magnetic susceptibility is defined as the ratio of magnetization M of the substance to the magnetic intenstiy

H of the magnetizing field in which it is placed. and is measured in A/m [ampere/metre]

The formula is given by $\chi$ = M /H------------(3)

WHERE $\chi$ is called susceptibility.

here H = B / $\mu$---------(4)

substituting for B = 5.95x10^-4 T

to know the value of $\mu$ in eqn (4) apply the formula as

B = $\mu$ n I Where n = 12500turns I = 0.0379A

by applying in the above formula we get

$\mu$ = 1,25 x 10^-6

knowing the values of $\mu$ and B    H is calculated and is given by

from eqn (4)

H = 5.95x10^-4 / 1.25x10^-6

H = 476 A/m (ampere/metre)

now we need to calculate the value of $\chi$

for this use the formula as

$\mu$r = $\mu$ / $\mu$0 = 1 + $\chi$----------(5)

we get

500 = (1.25 x 10^-6) / 4$\pi$ x 10^-7 = 1 + $\chi$

so $\chi$ = 496.6

since M and H have same units (A / m) the susceptibility has no units.

knowing the values of $\chi$ and H we substitute in eqn(3) and we get

$\chi$ = M / H

496.6 = M / 476

SO MAGNETIZATION M = 236.38 x 10³ (A/m)