Point of application of magnetic torque.

anmol21

1. The problem statement, all variables and given/known data

Hello,
When a current carrying loop (of any shape) is placed in a uniform magnetic field B , the field is given by MχB where M is the magnetic moment vector I*A where I Is current and A is the area vector. I want to know at which point or for which axis is the torque applied given by the above equation.

2. Relevant equations

ζ=MχB
M=I*A

3. The attempt at a solution

tiny-tim

hi anmol21! welcome to pf!

if B is uniform, the resultant force will be zero, so the effect is a pure torque

so the circuit will turn about the centre of mass of whatever the circuit is fixed to (or about whatever axis that is constrained to turn around)

anmol21

Thanks tiny-tim :)

So is the magnetic torque independent of the axis chosen ?

For example , Consider a sphere with a ring attached to it , with current I in the ring , in rotational equilibrium on a rough inclined plane. The magnetic field is vertically downwards and gravity is present. So we can equate the torque due to gravity and the magnetic torque at any point . So will be the magnetic torque be same even at the point of contact of the sphere with the ground or the centre of the sphere ?

tiny-tim

hi anmol21!

the magnetic torque of a uniform magnetic field on a circuit is a couple …

it is exactly the same as a couple (or "pure moment") in mechanics: it has the same moment about any point

"a system of forces with a resultant (a.k.a. net, or sum) moment but no resultant force"

"… the moment (torque) of a couple is independent of the reference point P: Any point will give the same moment. In other words, a torque vector, unlike any other moment vector, is a "free vector"."

both quotes from http://en.wikipedia.org/wiki/Couple_(mechanics)

anmol21

thanks a lot , that clears it up :)