How does the magnetic force affect torque on a current-carrying wire?

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SUMMARY

The discussion focuses on the effects of magnetic force on torque in a current-carrying wire, specifically analyzing segments QP, PR, and QR. The magnetic force on PR is directed into the plane with a magnitude of 0.8IB, while the force on QR is directed out of the plane, calculated as 0.8IB sin(arctan(0.8/0.6)). The net force on segment QR results in zero torque, as the opposing forces cancel each other out. The torque vector is confirmed to rotate outside the page, and the calculation of torque using the integral τ = ∫||r × dF|| yields a consistent result of 0.6IB/2.

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Homework Statement
The loop (negligible mass) in the figure is a right triangle in ##P##, and carries a current ##I = 500\,A##. A magnetic field of magnitude ##B=3.00\,T## and direction as shown in the figure fills the space.
a) Find the magnetic force (magnitude and direction) acting on each side of the loop.
b) Deduce the net force on the loop.
c) If the loop is pivoted about an axis lying along ##PR##, what is the magnitude of the net torque on the loop?
d) Will the loop rotate into or out of the plane?
Relevant Equations
.
Screenshot_2020-12-05_10-07-48.png

a) On QP, the force is zero.
On PR, the force is directed inside the plane, and its magnitude is ##0.8IB##.
On QR, the force is direction out of the plane, and its magnitude is ##1IB\sin(\arctan(0.8/0.6))=0.8IB##.
b) ##0.8IB-0.8IB=0##.
d) It will rotate outside of the page.
c) I know that the torque vector should be directed to the right because of d), but where is the magnetic force on QR applied to exactly? The centre?
I'm not really sure. Since the current is everywhere on QR, then should I look at every element of the wire and cross each force ##d\vec F## with the corresponding perpendicular distance vector from ##PR##, and integrate? Like ##\tau=\int||\vec r\times d\vec F||##?
 
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Using ##\tau=\int||\vec r\times d\vec F||##, I find that ##\tau=0.6IB/2##, which would be the same result of if I have considered the magnetic force to be acting on the centre of ##QR##, and took ##r=0.3=0.6/2##.
 

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