Force acting on a current carrying hollow conductor in a magnetic field

Thread Starter

sss2009

Joined Sep 24, 2019
2
It is known that when a current carrying conductor is placed within a magnetic field and current flows perpendicular to the magnetic field, a force is exerted on the conductor.

In figure 1, a square hollow conductor is shown in which DC current, I, is flowing perpendicular to the magnetic field, B. By Flemings left hand rule, the force acts downwards on the whole conductor.

What happens when the same current carrying conductor is subject to a magnetic field generated within? In figure 2, the same hollow square conductor is wrapped with an energized coil and magnetic field is generated within its cross section. DC Current is made to flow through the conductor in the same way as figure 1. Would a force be exerted perpendicular to each of the four sides of the square cross-section? forum pic.jpg
 

Glenn Holland

Joined Dec 26, 2014
665
Faraday's Law says the force is always at a right angle to the direction of the current flow.

This applies to all shapes of the conductor. For the coil wrapped around an an iron core, the flux forms a loop so there is a field within the coil and also on the outside.So there is no net force on the wire. The field wraps around a straight wire and there is no net force.
 

Thread Starter

sss2009

Joined Sep 24, 2019
2
Faraday's Law says the force is always at a right angle to the direction of the current flow.

This applies to all shapes of the conductor. For the coil wrapped around an an iron core, the flux forms a loop so there is a field within the coil and also on the outside.So there is no net force on the wire. The field wraps around a straight wire and there is no net force.
Dear Glenn,
Thank you for your reply. I just needed one small clarification.

The force acting on the conductor in Figure 1 is given by F = B I L, where L is the length of the conductor. This is my understanding from standard textbooks on the subject.forum pic2.jpg

What I really wanted to know was whether the same equation applies to each side of the square cross section in figure 2 (please see the revised figure attached)? Is that why there is no net force, since they cancel out each other?
 

Glenn Holland

Joined Dec 26, 2014
665
Dear Glenn,
Thank you for your reply. I just needed one small clarification.

The force acting on the conductor in Figure 1 is given by F = B I L, where L is the length of the conductor. This is my understanding from standard textbooks on the subject.View attachment 186837

What I really wanted to know was whether the same equation applies to each side of the square cross section in figure 2 (please see the revised figure attached)? Is that why there is no net force, since they cancel out each other?
Here's a good explanation of the interaction of current within any conductor and the resulting magnetic field commonly know as the Hall Effect:

https://en.wikipedia.org/wiki/Hall_effect

For a conductor placed in an external magnetic field, the current is diverted over to one side of the conductor. Or when current is flowing through a conductor, the resulting field is forced to the outside of the conductor. In that case the magnetic field resembles a tube surrounding the conductor.

The plasma in a lighting bolt is a particularly interesting example of the Hall Effect where the magnetic field surrounding the plasma produces a "Pinch Effect" that compresses the plasma into a "thread" of very high current density..
 
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