A 290 turn solenoid with a length of 21.0 cm and a radius of 1.80 cm carries a c

Question

A 290 turn solenoid with a length of 21.0 cm and a radius of 1.80 cm carries a current of 2.10 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. Incorrect: Your answer is incorrect. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. 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

Explanation / Answer

(a) using amperes law

mag field = u * current * number of turns / length =

= 4pix10^-7 * 290 * 2.10 / 0.21 =

= 0.003644 Tesla

(b) after 0.900 s, the current is 5.00 so

mag field = u * current * number of turns / length =

= 4pix10^-7 * 290 * 5.00 / 0.21 =

= 0.008677 Tesla

(c) area = pi r^2 = pi*0.018^2 = 0.001018 m^2

(d) change in mad flux = (final mag field - initial mag field) * area =

= (0.008677 - 0.003644) * 0.001018 = 5.123 x 10^-6 Wb

(e) emf = number of turns * change in flux / time =

= 4 * 5.123x10^-6 / 0.900 =

= 2.277 x 10^-5 Volts

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