In the figure (Figure 1) a conducting rod of length L = 38.0 cmmoves in a magnet

Question

In the figure (Figure 1) a conducting rod of length L = 38.0 cmmoves in a magnetic field B  of magnitude 0.420 T directed into the plane of the figure. The rod moves with speed v = 5.50 m/sin the direction shown.

http://postimg.org/image/zat4jxxfd/

Part A

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

Part B

Which point, a or b, is at higher potential?

Part C

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

Part D

What is the direction of the electric field within the rod?

Part E

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

Part F

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

Part G

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

Explanation / Answer

B = 0.42 T

L = 38 cm = 0.38 m

v = 5.5 m/s

(a) Potential difference = E = B L v = 0.42 * 0.38 * 5.5 = 0.9778 V

(b) From Lenzs Law , B is at higher potential

(c) E = v B = 5.5 * 0.42 = 2.31 V/m

(d) The electric field points from b to a,

(e) Point b has excess positive charge

(f) If the rod moves parallel to ab , Potential difference = E L = E*0 = 0

(g) Then , v and B will be same direction , Potential difference = 0

(e)

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