A long straight wire carrying a current of 2.12 A moves with a constant speed v
ID: 1483012 • Letter: A
Question
A long straight wire carrying a current of 2.12 A moves with a constant speed v to the right. A 120 turn circular coil of diameter 1.50 cm, and resistance of 3.25 µ, lies stationary in the same plane as the straight wire. At some initial time the wire is to the left of the coil at a distance d = 15.0 cm from its center. 4.00 s later, the wire has moved to the right of the coil and is at a distance d from the center of the coil. You may assume for simplicity that the magnetic field is uniform in the region of the coil.
(a) What is the direction of the induced current in the coil as the wire moves toward the coil, in the initial situation?
(clockwise) or (counterclockwise) or (no current) "multiple choice"
(b) What is the direction of the induced current in the coil as the wire moves away from coil, in the final situation?
(clockwise) or (counterclockwise) or (no current) "multiple choice"
(c) What is the magnitude of the average induced current in the coil over the 4.00 s interval?
?mA
Explanation / Answer
a) clockwise
b) counterclockwise)
c) Area of the loop, A = pi*d^2/4
= pi*0.015^2/4
= 1.77*10^-4 m^2
No of tunrs, N = 120
I = 2.12 A
t = 4 s
Let us assume magnetic filed through the loop
in the initial position
B1 = mue*I/(2*pi*d)
= 4*pi*10^-7*2.12/(2*pi*0.15)
= 2.83*10^-6 T (out of the page)
in the final position
B2 = mue*I/(2*pi*d)
= 4*pi*10^-7*2.12/(2*pi*0.15)
= 2.83*10^-6 T (in to the page)
Average induced emf = N*A*dB/dt
= 120*1.77*10^-4*2*2.83*10^-6/4
= 3*10^-8 volts
induced current = induecd emf/R
= 3*10^-8/(3.25*10^-6)
= 0.00923 A
= 9.23 mA <<<<<<<<------------------Answer