Torque equation of an electric motor

  • Thread starter fonz
  • Start date
  • #1
141
5

Main Question or Discussion Point

The example I am looking at in my text book 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
 

Answers and Replies

  • #2
1,506
17
I agree with you. Without seeing any diagram I would say the force on the side of the coil of length L will be BIL. The torque produced by the coil will be BILd where d is the separation of the sides of length L. Ld is the area of the coil.
If the coil is in a uniform field then the torque will be BIASinθ where θ is the angle between the field and the normal to the plane of the coil.
If the coil has N turns then the torqe is BNIASinθ
I don't understand what seems to be in your book !!!!!
hope this is some help
 

Related Threads for: Torque equation of an electric motor

  • Last Post
Replies
1
Views
1K
Replies
7
Views
27K
  • Last Post
Replies
2
Views
9K
  • Last Post
2
Replies
35
Views
5K
Replies
5
Views
771
  • Last Post
Replies
11
Views
868
  • Last Post
Replies
7
Views
11K
Replies
3
Views
2K
Top