A 310 turn solenoid with a length of 18.0 cm and a radius of 1.65 cm carries a c

Question

A 310 turn solenoid with a length of 18.0 cm and a radius of 1.65 cm carries a current of 2.25 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 310 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 310 turn solenoid.
T

(b) Calculate the magnetic field of the 310 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.

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

The force on the 2.1 kg (= m1) block is m1*g + k*x1 = 0 before the 315 g (=m2) mass is added.
After m2 is added the force on block m1 is (m1+m2)g + k(x1+x2) = 0.
So we have a system of two equation and two unknowns, k and x1:
m1*g + k*x1 = 0
(m1+m2)g + k(x1+x2) = 0
Solve for k and compute.

For a spring mass system the angular frequency w=(k/m)^(1/2)=2*pi/T, solve for T and compute.

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