A 290 turn solenoid with a length of 18.0 cm and a radius of 1.20 cm carries a c

Question

A 290 turn solenoid with a length of 18.0 cm and a radius of 1.20 cm carries a current of 1.80 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 290 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 290 turn solenoid.
T

(b) Calculate the magnetic field of the 290 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?

Explanation / Answer

a. Initial magnetic field inside the coil , B0 = 0nI where 0 is free space permiability 1.26 *10-6 NA-2

B0 = 7.67 * 10-4 NA-1

b. Final magnetic field , Bf = 18.27 * 10-4 NA-1

c. Area of 4 turn coil , A = *1.44 * 10-4 m2 = 4.52 *10-4 m2

d. Change in magnetic flux, = (Bf - B0) A = 10.6 *10-4 NA-1 * 4.52 *10-4 m2 = 4.79 *10-9 Nm2 A-1

f. Average induced emf = - 4.79 *10-9 Nm2 A-1/0.9 s = - 5.32 *10-9 Nm2 A-1 s-1

This is equal to instantaneous induced emf because the magnetic field increases steadility so at each moment d/dt is equal to /t .

g. Contribution to magnetic field due to current in 4 turn coil is neglected because it is small compared to the 290 coil as B is proportional to turns of coil.

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