Question

An air-filled solenoid has 360 turns of wire, a length of 20.0cm, and a cross- sectional area of 5.00cm^2. a) If the current is 5.00A, calculate the magnetic field in the solenoid. b) Calculate the self-inductance of the solenoid. c) If current increases at a rate of 50A per second what is the electromotive force across this solenoid?

Answer 1

Answer:

(a) 0.0113 T

(b) 4.07 x 10^-4 Henry

(c) 0.02 Volt

Explanation:

number of turns, N = 360

length, l = 20 cm = 0.2 m

Area of crossection, A = 5 cm^2 = 5 x 10^-4 m^2

i = 5 A

(a) The formula for the magnetic field of the solenoid

$B=\mu _{0}ni$

where, B be the strength of magnetic field, n be the number of turns per unit length, i be the current in the solenoid and μo is the absolute permeability of free space.

So, n = N / l = 360 / 0.2 = 1800 per metre

So, magnetic field strength

$B=4\times3.14\times10^{-7}\times1800\times5$

B = 0.0113 T

(b) Let L be the self inductance of the solenoid

$L =\frac {\mu _{0}N^{2}A}{l}$

Where, N be the total number of turns, A be the area of crossection

$L =\frac{4\times3.14\times10^{-7}\times360^{2}\times5\times10^{-4}{0.2}$

L = 4.07 x 10^-4 Henry

(c) di / dt = 50 A/s

The relation between the electromotive force and the rate of change of current is given by

$e = - L\times \frac{di}{dt}$

e = - 4.07 x 10^-4 x 50 = - 0.02 Volt

Negative sign shows the direction of electro motive force.