Answer the following questions with regard to the diagram above. If the first is

Question

Answer the following questions with regard to the diagram above. If the first is T and the rest F, enter TFFFF.

A) As the rod AB moves to the right, the magnitude of the flux through the surface ABCD decreases.
B) The magnetic field due to the INDUCED current points within the loop ABCD into the plane of the paper.
C) The induced current flows clockwise.
D) In the absence of friction between rod and rail, the rod will continue to move towards the right indefinitely even if no external force is applied to keep it in motion.
E) The agent moving the rod to the right is doing negative work on the rod

Explanation / Answer

given that

length of rod d = 1.40 m

mass m = 0.20 kg

flux density B= 350 mWb/m^2 = 0.35 Wb/m^2

velocity v = 6.10 m/s.

Induced emf

e = delta phi / delta t

where delta phi / delta t is the total rate of change flux by the movement of the rod.

e = delta phi / delta t = B * d * (delta x / delta t)

e = B * d * v        (delta x / delta t = velocity v)

e = 0.35 * 1.4 * 6.10 = 2.98 V

(A) True . As the rod AB moves to the right, the flux through the surface ABCD decreases.

(B) false . as we know that the induce current direction will be such that it will oppose in change in flux of loop so the direction of induced magnetic feild will be out of the plane of the paper upper side.

(C) False . The induced current flows counter-clockwise.

(D) True . In the absence of friction between rod and rail, the rod will continue to move towards the right indefinitely even if no external force is applied to keep it in motion.

(E) False .The agent moving the rod to the right is doing positive work on the rod because force and displacement both are in same direction

so the answer is TTFTF

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