Derivation Of Torque On Current Loop Due To Uniform Magnetic Field

[Aurelius120]

Homework Statement

Derive an expression for torque due to uniform magnetic field in a loop

Relevant Equations

$$\tau=BINA\sin \theta$$

I can derive it for a circular loop:
$$dF=BI\sin\phi\ dl=BIr\sin\phi\ d\phi$$
Torque on quarter circle when field is parallel to plane of loop=$$\tau=\int^{(\pi/2)}_0 BI \ dl \sin\phi (r\sin\phi)$$$$=\int^{(\pi/2)}_0 BIr^2 \sin^2\phi\ d\phi$$
Net torque=$4\tau=BIA$
If magnetic field is at any other angle, only its parallel component will exert torque=$BIA\sin\theta$
I know the derivation for rectangular loop(length=l, breadth=b).
Force on each arm =$IbB$
Torque=$2IbB\frac{a}{2}=BIA$
If magnetic field is at any other angle, only its parallel component will exert torque=$BIA\sin\theta$
Is there a general derivation for such cases that holds for (at least the most common shapes)? Will it be too advanced for my level?

 

[Orodruin]

Write down the integral expression for the torque as a circulation integral around the loop. Then apply a suitable integral theorem.

 

[Aurelius120]

Write down the integral expression for the torque as a circulation integral around the loop. Then apply a suitable integral theorem.

Ok that seems advanced for me

 

[Orodruin]

Then prove it for a square loop, then approximate the full loop by a series of several ever smaller square loops (the torques from sides shared by two squares will cancel out since the current runs in opposite directions - leaving only the outer loop). However, this is essentially proving the integral theorem.

 

[Delta2]

Write down the integral expression for the torque as a circulation integral around the loop. Then apply a suitable integral theorem.

Hasn't he done almost this in the OP, the only thing that changes for a generic loop is that r depends on phi, right?

EDIT: OH I think in the OP he omits the other vector constituent of the torque ($BIdl\sin\phi (r\cos\phi)$ which sums to zero for a circular loop but doesnt sum to zero for a generic loop.)

EDIT2: On second thought that integral is zero for any loop shape. This thread triggered an interesting not so hard math problem:
If $r(\phi)$ is a positive function with period $2\pi$ prove that $$\int_0^{2\pi} r^2(\phi)\sin(\phi)\cos(\phi)d\phi=0$$ and $$\int_0^{2\pi} r^2(\phi)=2A$$ where A the area of the closed loop described by $r=r(\phi)$.

EDIT3: On ... third thought this problem is not so simple after all if one wants a mathematically rigorous proof for any closed loop but ok I tried my best, I haven't read any derivation from a book from this I ll check Jackson to see if it has it.