A 20-turn wire coil in the shape of a rectangle. 0.25 m. by 0.15 m. has a resist
ID: 2010196 • Letter: A
Question
A 20-turn wire coil in the shape of a rectangle. 0.25 m. by 0.15 m. has a resistance of 5.0 Ohm. In position shown above, the loop is in a uniform magnetic fields B of 0.20 T. The field is directed out of the page. perpendicular to the plane of the loop. The loop is pulled to the right at a constant velocity, reaching position 2 m 0.50 s. where B is equal to zero. Calculate the average emf. induced in. the 20-turn coil during this period. Calculate the magnitude of the current induced in the 20-turn coil and stale its direction. Calculate the power dissipated in the 20-turr. coil. Calculate the magnitude of the average force necessary to remove the 20-turn. coil from the magnetic field. Identical wire is used to add 20 more turns of wire to the original coil. How docs this affect the current in the coil Justify veer answer.Explanation / Answer
Given that number turns in coil is N = 20 Area of the rectangle is A = 0.25 m * 0.15m = 0.037 m^2 magnetic field is B = 0.20 T time t = 0.50 s a ) Averge emf during this period is e = -N d / dt = N * dB A / d t = 20 * ( 0.20 T - 0 ) * 0.037 m^2 / 0.50s = 0.29V b ) Magnitude of current is i = e / R = 0.29 V / 5.0 = 0.059A c ) Power dissipated in the coil is P = V * I = 0.29 V * 0.059 A = 17.1*10^-3W d ) Magnetic force F = i L B = 0.059 A * 0.25 m * 0.20 T = 2.95 *10^-3 N e ) When number of turns increased emf is change and the current will changeRelated Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.