N neutral metal rod of length 0.45m slides horizontally to the left at a constan

Question

N neutral metal rod of length 0.45m slides horizontally to the left at a constant speed of 7 m/s on frictionless insulating rails through a region of uniform magnetic field of magnitude 0.4 tesla, directed out of the page as shown. Before answering the following questions, draw a diagram showing the polarixation of the rod, and the direction of the Coulomb electric field inside the rod.

A) Which of the following statements is true?

1) The top of the moving rod is positive.

2)The top of the moving rod is negative.

3) The right side of the moving rod is positive.

4) The right side of the moving rod is negative.

5) The moving rod is not polarized.

B) After the initial transient, what is the magnitude of the net force on a mobile electron inside the rod?

C) What is the magnitude of the electric force on a mobile electron inside the rod?

D) What is the magnitude of the magnetic force on a mobile electron inside the rod?

E) What is the magnitude of the potential differance across the rod?

F) In what direction must you exert a force to keep the rod moving at a constant speed?

Explanation / Answer

I just finished working a problem similar to this so I'll try and help out with the ones I feel confident with. A) If the magnetic force is out of the page, then the induced magnetic field created by the moving rod should point into the page. Using the RHR pointing the thumb into the page the current will move Clockwise resulting in the top of the rod being positively charged. B) C) Magnitude of Electric Force F = qE D) Magnitude of Force of B = qvBsin? So F = -1.6 x 10^-19 x 7 m/s x 0.4 T = 4.48 x 10^-19 N E) Using Faraday's Law you can obtain the equation Emf = Blv So the potential difference would be 0.4 T x 0.45m x 7 m/s = 1.26 V F) I assume the direction of the force must be west to keep the rod at constant speed If you have the answers to the problem i'd like to know what the correct ones are. Also, if I had the answers for B/C I could figure out how to obtain it.

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