A solenoid 20.0 cm long and with a cross-sectional area of 0.500 cm^2 contains 3

Question

A solenoid 20.0 cm long and with a cross-sectional area of 0.500 cm^2 contains 395 turns of wire and carries a current of 81.5 A. (a) Calculate the magnetic field in the solenoid. (b) Calculate the energy density in the magnetic field if the solenoid is filled with air. (c) Calculate the total energy contained in the coil's magnetic field (assume the field is uniform). (d) Calculate the inductance of the solenoid. An inductor with an inductance of 2.40 H and a resistance of 7.00 Ohm is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance. (a) Find the initial rate of increase of current in the circuit. (b) Find the rate of increase of current at the instant when the current is 0.500 A. (c) Find the current 0.250 s after the circuit is closed. (d) Find the final steady state current.

Explanation / Answer

2)

i = 81.5 A

l = 20 cm = 0.2 m

Cross-sectional area A = 0.5 cm2 = 0.5 x 10-4 m2

n = 395/0.20 = 1975 turns per metre

permeability of free space = 0 =1.2566 × 10-6 H/m

Magnetic field: B = 0 n i = 1.2566 × 10-6 x 1975 x 81.5 = 0.202265 = 0.2023 T

Inductance: L = 0 n2 A l = 1.2566 × 10-6 x 19752 x 0.5 x 10-4 x 0.2 = 0.00004902 H = 0.4902 x 10-4 H

Energy stored in the solenoid: U = (1/2) L i2 = (1/2) x 0.4902 x 10-4 x 81.52 = 0.1628 J
Energy density of magnetic field: uB = U/volume = U/(A x l) = 0.1628 / (0.5 x 10-4 x 0.2) = 16280.2 J/ m3

3)

Instantaneous Current It= (Vs/R) (1-e-Rt/L)

differentiating w r t 'time', dIt/dt= (Vs/L) e-Rt/L

(a) Initial rate of increase of current ( t = 0) , (dIt/dt)t=0 = (Vs/L) e0 = (6/2.4) = 2.5 A

(b) Rate of increase of current at I = 0.5 A,

It= (Vs/R) (1-e-Rt/L) rearranging this equation gives, e-Rt/L = (1- RIt/Vs)

from above (a), dIt/dt= (Vs/L) e-Rt/L   substituting e-Rt/L = (1- RIt/Vs), gives

dIt/dt= (Vs/L) (1- RIt/Vs)

here  It = 0.5 A, thus, (dIt/dt)t = 0 = (6/2.4) (1-(7x0.5)/6) = 1.041 A

c) the circuit current at t = 0.25 s after the switch is closed. is

It= (Vs/R) (1-e-Rt/L) = (6/7) (1- e-(7x0.25)/2.4) = 0.4437 A

d) Steady state current I = Vs/R = 6/7 = 0.8571 A

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