In the figure, a conducting rod of length L = 27.0 cm moves in a magnetic field

Question

In the figure, a conducting rod of length L = 27.0 cm moves in a magnetic field B of magnitude 0.360 T directed into the plane of the figure. The rod moves with speed v = 7.00 m/s in the direction shown. (Figure 1)

1 What is the direction of the electric field in the rod?

2 When the charges in the rod are in equilibrium, what is the magnitude E of the electric field within the rod?

3 Which point, a or b, has a higher potential?

4 What is the magnitude Vba of the potential difference between the ends of the rod?

5 What is the magnitude E of the motional emf induced in the rod?

Explanation / Answer

Here ,

L = 27 cm = 0.27 m

magnetic field , B = 0.360 T

v = 7 m/s

1)

as the magentic force on the particles is given as

F = q * V X B

the force on the positive charges is to right

hence , electric field is to the right

2)

for the electric field is E

E = v * B

E = 7 * 0.360 N/C

E = 2.52 N/C

the electric field E is 2.52 N/C

3)

the point b is at higher potential

4)

potential difference , Vba = E * L

potential difference , Vba = 2.52 * 0.27

potential difference , Vba = 0.68 V

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