- #1
fonz
- 151
- 5
The example I am looking at in my textbook starts by considering and area of the rotor surface of width w and length L.
Then the axial current flowing in the width w is equal to I=wA which is exposed to a radial flux density B
So from the Lorentz force F=IBxL the equation becomes
F= wABxL
so the force per unit area is F/wL which becomes:
F=BA
Then to obtain the torque the force per area is multiplied by the entire area of the rotor (2∏rL) then multplied by the radius of the rotor
So the overall torque equation becomes:
T=BA x 2∏rL x r
What doesn't make sense is how can the current be equal to wA? by the Lorentz equation the force on a current carrying conductor is IBxL so the width and area of the conductor carrying the current I does not matter? so why does it apply here?
Regards
Dan
Then the axial current flowing in the width w is equal to I=wA which is exposed to a radial flux density B
So from the Lorentz force F=IBxL the equation becomes
F= wABxL
so the force per unit area is F/wL which becomes:
F=BA
Then to obtain the torque the force per area is multiplied by the entire area of the rotor (2∏rL) then multplied by the radius of the rotor
So the overall torque equation becomes:
T=BA x 2∏rL x r
What doesn't make sense is how can the current be equal to wA? by the Lorentz equation the force on a current carrying conductor is IBxL so the width and area of the conductor carrying the current I does not matter? so why does it apply here?
Regards
Dan