A 0.87-m aluminum bar is held with its length parallel to the east-west directio

Question

A 0.87-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 18 m/s, and the emf induced across its length is 6.8 x 10-4 V. Assuming 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, and (b) state whether the east end or the west end of the bar is positive.

Explanation / Answer

(a)

Whenever EMF is induced by means of magnetism, one should use Faraday's law, which states:
EMF = -d/dt
To use it in this problem, you have to adjust formula.
=B*S*cosa=B*l*x*cosa
If inserted into the first equation, you get
EMF = -d/dt=-d(B*l*x*cosa)/dt=-B*l*cosa*dx/dt... where
- magnetic flux
B - magnetic field induction
a - angle between the normal of the plane (in which the bar falls) and the magnetic field induction B
l - bar's length
v- speed of te bar
From initial conditions you know l, v, EMF and a=0°
Forget about "-" sign. It gives information about direction, which for now is irrelevant.
B=EMF/(l*v*cosa)=6.84 x 10^-4/(0.87*18*1)= 4.367x10^-5= 43.6T

(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.

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