In the figure, a conducting rod with length L = 29.0 cmmoves in a magnetic field

Question

In the figure, a conducting rod with length L = 29.0 cmmoves in a magnetic field B  of magnitude 0.430 Tdirected into the plane of the figure. The rod moves with speed v = 7.00 m/s in the direction shown. (Figure 1)

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

Express your answer in volts per meter to at least three significant figures.

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

Express your answer in volts to at least three significant figures.

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

Express your answer in volts to at least three significant figures.

Explanation / Answer

a) The electric field points from b to a, resisting further charge separation. As we showed bellow, E = vB = 3.014 V/m

b) The conducting rod has mobile charges of charge q. These charges feel a magnetic force from passing through the magnetic field FB = |FB| = |qv × B| = qvB , because v and B are perpendicular

This force pushes the charges from one end of the rod to the other which causes Q = n|q| to build up on one side of the rod (b) and Q to be exposed on the other side (a). The charge separation increases until the electric force FE = qE (8) balances the magnetic force

qE = qvB (9) E = vB

Because the electric field in the bar is constant, the electric potential difference Vab is given by Vab = EL = vBL = 7.00 m/s · 0.430 T · 0.290 m = 0.873mV

c) E = B/ t = ( / t)(L W B) = L B W t = LBv = 0.290 m · 0.430 T · 7 m/s = 0.873mV

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