A 300-turn solenoid with a length of 20.0 cm and a radiusof 1.50 cm carries a current of 2.00 A. A second coil of fourturns is wrapped tightly around this solenoid, so it can be consideredto have the same radius as the solenoid. The currentin the 300-turn solenoid increases steadily to 5.00 A in 0.900 s.(a) Use Ampère’s law to calculate the initial magnetic field inthe middle of the 300-turn solenoid. (b) Calculate the magneticfield of the 300-turn solenoid after 0.900 s. (c) Calculatethe area of the 4-turn coil. (d) Calculate the change in themagnetic flux through the 4-turn coil during the same period.(e) Calculate the average induced emf in the 4-turn coil. Isit equal to the instantaneous induced emf? Explain. (f ) Whycould contributions to the magnetic field by the current in the4-turn coil be neglected in this calculation?
A 300-turn solenoid with a length of 20.0 cm and a radius
of 1.50 cm carries a current of 2.00 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 300-turn solenoid increases steadily to 5.00 A in 0.900 s.
(a) Use Ampère’s law to calculate the initial magnetic field in
the middle of the 300-turn solenoid. (b) Calculate the magnetic
field of the 300-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. (f ) Why
could contributions to the magnetic field by the current in the
4-turn coil be neglected in this calculation?
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