I'm assuming the physical set up looks like this:
![Van Gogh]()
Assuming the current, II, is moving CCW as indicated, then the effect of the B field, again acting in the direction shown, will be to create a torque vec tau→τ that spins the loop about it's axis as shown. In this set-up, from the right hand rule, the force vectors on the RHS of the blue vertical axis will act out of the page, and vice versa:
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So we do that calculus thing and linearise a small element of the circular loop:
![Cezanne]()
The force vec F →F on a current-carrying wire of length and orientation vec L→L, due to a magnetic field vec B→B (BTW this is all derived from the Lorentz Force Law), is:
vec F = I (vec L xx vec B) →F=I(→L×→B)
From the drawing: abs(vec L) = R d theta∣∣∣→L∣∣∣=Rdθ
So, again from the right hand rule, the force d vec Fd→F on this small element is:
d vec F = I abs(vec L) abs(vec B) sin (pi/2 - theta) \ hat z, i.e. acting out of the screen/page.
= I \ R \ d theta \ B \ cos theta \ hat z
The associated torque d vec tau is:
d vec tau = vec r xx d vec F, where vec r is the small element's position vector relative to its axis of rotation .
In this case: vec r = R cos theta \ hat x
And so:
d vec tau = R cos theta \ hat x xx I \ R \ d theta \ B \ cos theta \ hat z
= - IBR^2 cos^2 theta \ d theta \ hat y
implies vec tau = - IBR^2 int_0^(2 pi) cos^2 theta \ d theta \ hat y
Simple maths sub and...
= - pi IBR^2 \ hat y
That means that:
implies abs(vec tau) = pi IBR^2
= pi (28 xx 10^(-3))( 0.9) (1.3/2)^2 = 0.033 Nm
