A long, straight wire carrying a current of 3.15 A moves with a constant speed v

Question

A long, straight wire carrying a current of 3.15 A moves with a constant speed v to the right. A 5-turn circular coil of diameter 1.25 cm, and resistance of 3.25 , lies stationary in the same plane as the straight wire. At some initial time, the wire is at a distance d = 11.0 cm from the center of the coil. 4.55 s later, the wire is at a distance 2d from the center of the coil. What is the magnitude and direction of the induced current in the coil? Note that while the magnetic field varies over the diameter of the coil, it is very small and we will disregard this variation. 0.475 magnitudeWhat is the change in magnetic flux inside the coil due to the straight wire? How does Faraday's law relate this change in flux to the induced emf? mA | counterclockwise direction 2d Initial situation Final situation

Explanation / Answer

Initial magnetic field at the center of the loop,

B1 = mue*I/(2*pi*d)

= 4*pi*10^-7*3.15/(2*pi*0.11)

= 5.73*10^-6 T

Final magnetic field at the center of the loop,

B2 = mue*I/(2*pi*2*d)

= 4*pi*10^-7*3.15/(2*pi*2*0.11)

= 2.86*10^-6 T

Induced emf = change in magnetic flux/time taken

= N*A*(B1 - B2)/t

= 5*pi*(0.0125/2)^2*(5.73 - 2.86)10^-6/4.55

= 3.87*10^-10 V

induced current = induced emf/R

= 3.87*10^-10/3.25

= 1.19*10^-10 A

= 1.19*1^-7 mA

direction : counterclockwise

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