Question

In the figure, a conducting rod of length *L* = 33.0 cm
moves in a magnetic field *B*? of magnitude 0.550 T directed
into the plane of the figure. The rod moves with speed *v* =
4.00 m/s in the direction shown.

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

2.What is the magnitude *V**b**a* of the
potential difference between the ends of the rod?

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

Answer #1

1)

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 = 4*0.550 = 2.2V

2)

emf V = E*d

d = 33cm = 0.33m

V = 2.2*0.33

V = 0.726V

3)

E = v*B

E = 4*0.550 = 2.2V

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