20.3 A short circular coil with 50 turns has a radius of 3 cm and a resistance o
ID: 2032493 • Letter: 2
Question
20.3 A short circular coil with 50 turns has a radius of 3 cm and a resistance of 1.7 S2. The coil is placed is a 2 T magnetic field with the plane of the coil perpendicular to the magnetic field. If the magnetic field goes to zero in 0.17 s, what is the induced voltage in the coil? b) What is the induced current in the coil? Include the direction of the current going around the coil. c) What is the induced magnetic field at the center of the coil? d) Calculate the ratio of the original magnetic field divided by the induced magnetic field. B=2TExplanation / Answer
(a) Induced voltage V = n*?B*A/?t
where A= area of loop is m^2
?B = change in field
?t = time taken
n= number of turns
So, V= 50 * (2-0) * pi*r^2 / 0.17 = 100*2.827*10^-3 / 0.17 = 1.663 V .
(b) Induced current = V/R = 1.663/1.7 = 0.978 A. as the initial field is decreasing is reduced, the current is such as to increase this field. So the direction is anticlockwise as seen from top
(c)Induced magnetic field at the center of the coil is B= n*?0*I / 2*r = 50*4*pi*10^-7*0.978 / 2*0.03 = 1.024*10^-3 T
(d) ratio of original to induced magnetic field is 2/1.024*10^-3 = 1952.8
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.