In the figure (Figure 1) a conducting rod of length L = 33.0 cm moves in a magne

Question

In the figure (Figure 1) a conducting rod of length L = 33.0 cm moves in a magnetic field B  of magnitude 0.490 T directed into the plane of the figure. The rod moves with speed v = 5.40 m/s in the direction shown.

What is the potential difference between the ends of the rod?

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

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

What is the potential difference across the rod if it moves parallel to ab?

What is the potential difference across the rod if it moves directly out of the page?

Explanation / Answer

Q1:

The potential difference between the ends of the rod is nothing but the emf:

V = BLv

V = 0.49T*33*10-2m*5.4m/s = 0.8732 V

Q2:

The magnitude of electric field is

E = V/d = V/L

E = 0.8732/33*10-2 = 2.646 N/C

Q3:

Excess of positive charge willl be at the point b as it possess higher potential

Q4:

The potential difference across the rod if it moves parallel to ab is zero, as the length of the rod decreases and tends to zero, therefore

V = 0

Q5:

When the rod moves directly out of the page the angle becomes 180o and the value will be

V = BLv sin 180o = 0

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