The sewer pipe solenoid you used in lab to measure the earth’s magnetic field ha

ID: 1461378 • Letter: T

Question

The sewer pipe solenoid you used in lab to measure the earth’s magnetic field had 79 turns wound on 79cm. It’s diameter is 4in (1in = 2.54cm).

(a)[3 pt(s) ]Compute the self-inductance.

(b)[2 pt(s) ]Compute the magnetic energy (ignoring the earth’s field) if 0.25A flows through the solenoid.

(c)[2 pt(s) ]Compute the magnetic energy density by dividing the energy by the volume of the solenoid.

(d)[1 pt(s) ]Compare the result in (c) with the result found from the formula for magnetic energy density.

diameter, d = 4 in

= 4*2.54 cm

= 10.16 cm

= 0.1016 m

cross sectional area, A = pi*d^2/4

= pi*0.1016^2/4

= 8.11*10^-3 m^2

length of solenoid, l = 79 cm = 0.79 m

no of turns, N = 79

a) self inductance, L = mue*N^2*A/l

= 4*pi*10^-7*79^2*8.11*10^-3/0.79

= 8.05*10^-5 H

b) magnetic energy, U = 0.5*L*I^2

= 0.5*8.05*10^-5*0.25^2

= 2.51*10^-6 J

c) magnetic energy density, u = U/V

= U/(A*l)

= 2.51*10^-6/(8.11*10^-3*0.79)

= 3.92*10^-4 J/m^3

d) magnetic filed inside solenoid, B = mue*N*I/L

= 4*pi*10^-7*79*0.25/0.79

= 3.14*10^-5 T

magnetic energy density, u = B^2/(2*mue)

= (3.14*10^-5)^2/(2*4*pi*10^-7)

= 3.92*10^-4 J/m^2

both are same.

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