In the figure, a conducting rod of length L = 31.0 cmmoves in a magnetic field B

Question

In the figure, a conducting rod of length L = 31.0 cmmoves in a magnetic field B  of magnitude 0.550 Tdirected into the plane of the figure. The rod moves with speed v = 5.00 m/s in the direction shown.

A)When the charges in the rod are in equilibrium, which point, a or b, has an excess of positive charge?

B)What is the direction of the electric field in the rod?

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

D)Which point, a or b, has a higher potential?

E)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.

F)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) Point B

b) From b towards a

c) when the rod is in motion charges experience magnetic force in presence of magnetic field and
according to Force(magnetic) = q*[cross product of velocity and field]
if velocity is on -ve y-axis and field inside the plane then by Right hand Screw rule positive charges experience force towards point P and negative charges accumulate at point A and hence naturally their is attractive coulombic force between charges so in steady condition these forces balance each other and Hence
q*v*B = q*E
E = v*B
E = 5*0.55 = 2.75 V/m

d) point b has higher potential

e) emf V = E*d
d = 31 cm = 0.31 m
V = 2.75*0.31
V = 0.85 V

f) same as part c)

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