A 330 turn solenoid with a length of 22.0 cm and a radius of 1.70 cm carries a c

Question

A 330 turn solenoid with a length of 22.0 cm and a radius of 1.70 cm carries a current of 1.70 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 330 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 330 turn solenoid.


(b) Calculate the magnetic field of the 330 turn solenoid after 0.900 s.

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


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


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


Is it equal to the instantaneous induced emf? Explain.

This answer has not been graded yet.

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

Explanation / Answer

given data
N = 330

L = 0.22 m

r = 0.017 m

i = 1.7 A

a) B1 = mue*N*i1/L = 3.203*10^-3 T

b)B2 = mue*N*i2/L = 9.42*10^-3 T

c) A = pi*r^2 = 9.07*10^-4 m^2

d) change in flux = A*(B2-B1) = 5.6*10^-6 weber

e) emf = change in flux/time = 6.2*10^-6 volts

no it is avearge induced emf.

f)because the induced emf is very small. so, the induced current very small.

the magnetic filed produced by this small current can be neglected.

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