A 0.82-m aluminum bar is held with its length parallel to the east-west directio

Question

A 0.82-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar hits the river below, its speed is 20 m/s, and the emf induced across its length is 6.60 10-4 V. Assume the horizontal component of the earth's magnetic field at the location of the bar points directly north. (a) Determine the magnitude of the horizontal component of the earth's magnetic field. T (b) State whether the east end or the west end of the bar is positive. East end of the bar is positive. West end of the bar is positive.

Explanation / Answer

This is called motional EMF. In simplest terms, if velocity v, length L and magnetic field B are mutually perpendicular, EMF = vLB. Since EMF is not a vector we can't refer to its direction, but we can determine the direction of the induced E field. From the ref., E = v X B, where X is the vector cross product. Then EMF = ELcos(theta) where theta is the angle between E and L (i.e., EMF = E dot L). (Note that L is bidirectional, so we can keep theta between 0 and 90 deg.)
A. Here we use the scalar equation since only the horizontal component is asked for.
B(hor) = EMF/(vL)

= 6.60 10-4 V /(0.82*20)

= 4.024E-5 T

B. Here we only need to know whether E has an east or a west component, so we use the right-hand rule with E = v X B. Since v points down and B has a north component, E has an east component. So answer is  East end of the bar is positive.

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