Question

A 320 turn solenoid with a length of 19.0 cm and a radius of 1.20 cm carries a current of 1.90 A. A second coil of four turns is wrapped tightly around this solenoid, so it can be considered to have the same radius as the solenoid. The current in the 320 turn solenoid increases steadily to 5.00 A in 0.900 s.

(a) Use Ampere's law to calculate the initial magnetic field in the middle of the 320 turn solenoid.
T

(b) Calculate the magnetic field of the 320 turn solenoid after 0.900 s.
T


(c) Calculate the area of the 4-turn coil.
m²

(d) Calculate the change in the magnetic flux through the 4-turn coil during the same period.
Wb

(e) Calculate the average induced emf in the 4-turn coil.
V

Is it equal to the instantaneous induced emf? Explain.

(f) Why could contributions to the magnetic field by the current in the 4-turn coil be neglected in this calculation?

Answer 1

consider a rectangular loop of length a and width b with half inside the solenoid

number of turns inside the loop = (N/L)*a

current through the loop Iloop = (N/L)*a*I

from amperes law

B*a = uo*Iloop

B*a = uo*(N/L)*a*I

B = uo*(N/L)*I

magnetic field B1 = 4*pi*10^-7*(320/0.19)*1.9 = 0.00402 T

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(b)

magnetic field B2 = 4*pi*10^-7*(320/0.19)*5 = 0.01058 T

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(c)

area of coil A = pi*r^2 = pi*0.012^2 = 4.52*10^-4 m^2

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(d)

change in magnetic flux = 4*A*(B2-B1) = 4*4.52*10^-4*(0.01058 - 0.00402) = 1.186*10^-5

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(e)

average enduced emf = rate of change in flux

average enduced emf = 1.186*10^-5/0.9 = 1.318*10^-5 V

equal to instantaneous

the current in the solenoid increases steadily

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f)

the induced emf is very small , the induced current in the 4 turn coil is very small

so the magnetic field due to current in 4 turn coil can be neglected