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Point of application of magnetic torque. 
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#1
Jul312, 02:41 PM

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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 


#2
Jul312, 05:52 PM

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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) 


#3
Jul412, 04:45 AM

P: 3

Thanks tinytim :)
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 ? 


#4
Jul412, 04:57 AM

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Point of application of magnetic torque.
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"both quotes from http://en.wikipedia.org/wiki/Couple_(mechanics) 


#5
Jul512, 05:18 AM

P: 3

thanks a lot , that clears it up :)



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