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Electricity and Magnetism

  • Thread starter garytse86
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1) A horizontal metal rod of 1.5m long is aligned in and East-West direction and dropped from rest from the top of a high building. Calculate the e.m.f. induced across the falling rod 2.5s after release. The horizontal component of Earths magnetic field is 2x10^5 T.

Surely you need the area perpendicular to field lines, which would be the area of a circle. so you need the radius of the rod?

2) Explain why positive paint drops are attracted to the metal object. This is about an electrostatic paint sprayer, and the metal object is earthed.
 

Answers and Replies

OlderDan
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garytse86 said:
1) A horizontal metal rod of 1.5m long is aligned in and East-West direction and dropped from rest from the top of a high building. Calculate the e.m.f. induced across the falling rod 2.5s after release. The horizontal component of Earths magnetic field is 2x10^5 T.

Surely you need the area perpendicular to field lines, which would be the area of a circle. so you need the radius of the rod?

2) Explain why positive paint drops are attracted to the metal object. This is about an electrostatic paint sprayer, and the metal object is earthed.
You don't need area for the first one. The rod is horizontal. Movement through the earth's magnetic field will create a magnetic force on the charges in the rod proportional to the field strength and rod velocity. Charges will flow to the ends of the rod until an electric field is created that offsets the magnetic force. This is somewhat similar to the problem of a charge fired at some velocity into a crossed electric and magnetic field and finding the velocity that results in zero deflection. Once you have the electric field strength needed for zero net force on any remaining charges, you can calculate the potential difference between the ends of the rod.

For the second, think about what happens when any charge is brought near a conducting plane, expecially one that is grounded (or earthed).
 
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Thanks. I got so far:

F= Bqv

v= u + at, so v = 24.5 m/s

F = 2 x 10^-5 x 24.5 x q
F = 4.9 x 10^-5 x q


Electric field strength = F/q
E = 4.9 x 10^-5
EMF = 4.9 x 10^-5 x distance travelled.

v^2 = u^2 + 2as
s = 30.625.

EMF = 1.5 x 10^-3 V.

Is this right?
 
OlderDan
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garytse86 said:
Thanks. I got so far:

F= Bqv

v= u + at, so v = 24.5 m/s

F = 2 x 10^-5 x 24.5 x q
F = 4.9 x 10^-5 x q


Electric field strength = F/q
E = 4.9 x 10^-5
EMF = 4.9 x 10^-5 x distance travelled.

v^2 = u^2 + 2as
s = 30.625.

EMF = 1.5 x 10^-3 V.

Is this right?
The distance involved in the emf calculation is not the distance the rod travels. That distance is perpendicualr to the direction of the electric field. The electric field is parallel to the length of the rod, and the induced emf is the potential difference between the ends of the rod.
 
308
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oh yeah so its EMF = 4.9 x 10^-5 x 1.5 = 7.4x10^-5V?

Thanks
 
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In the (1) part , the emf induced is known as "motional emf" . When the rod is falling down with velocity V , the electrons inside the rod start moving opposite to the direction of [itex]V x B[/itex] , as a result electric field is produced within the rod .With time, this electric field opposes further movement of electrons and thus an equilibrium is produced ,

At this equilibrium , emf induced = Blv
 

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