Lorentz Force in conductive beam

[diemilio]

It is known that in a current carrying wire, exposed to a magnetic field, the wire will experience a force equal to the product of the field, the current and the length of the wire (where the direction of the force is orthogonal to both the field and the direction along the length of the beam). F = B*I*L

Now, in a rectangular beam of length L, width w, and height h (where w<<h<<L), again carrying a current and in the presence of a magnetic field, how is the Lorentz force expressed? is the force acting as a distributed load all along the beam? is this integrated force modeled as a point load at the center of the beam? Is it acting as a body load? a surface load? Could someone help me understand this??

Thanks in advance,

diemilio

 

[elect_eng]

It is known that in a current carrying wire, exposed to a magnetic field, the wire will experience a force equal to the product of the field, the current and the length of the wire (where the direction of the force is orthogonal to both the field and the direction along the length of the beam). F = B*I*L

Now, in a rectangular beam of length L, width w, and height h (where w<<h<<L), again carrying a current and in the presence of a magnetic field, how is the Lorentz force expressed? is the force acting as a distributed load all along the beam? is this integrated force modeled as a point load at the center of the beam? Is it acting as a body load? a surface load? Could someone help me understand this??

Thanks in advance,

diemilio

In such a case the force is distributed, and it would be important to know the current density (current per unit area) in the beam. For a low frequency, or direct current, you may be able to approximate the current density distribution as constant across the beam, but it really depends on the material properties and how the voltage is applied to the ends of the beam.

In general the force would be expressed as an integral over the volume of the beam. The Lorenz force is really force per unit length, and the surface integral is taken over the cross section in terms of current density J, rather than current I.

 

[diemilio]

Nice! Got it! Thank you so much for the quick reply!

diemilio