Current carrying coil - torque on a circular one

[schmave]

Hello,

Could someone please either verify my proof or find the error in it - its driving me nuts!

if Torque (T) = Fd

but F = BIlsin theta and F changes as you move around the circle

1. therefore to find F on the semi-circle:

$$\int$$ between 0 and pi of BIlsin theta d theta = 2BIl ?

2. arc length 'l' of the semi circle is l = r phi -> l = r.pi

3. therefore F = 2BIrpi

4. into T= Fd -> T = 2BIrpi . 2r

5. therefore T = 4 BI. pi r squared

6. hence T = 4 BIA where A is the area of the circle?

please! say this is wrong, every time i look at it i wish it was becuase T = BIA for all planar shapes is just so much more simple :)

is faraday sufficiently spinning in his grave yet ?

 

[Vailanor]

!Thanks for your help.Your proof is incorrect. The torque formula you are using, Torque (T) = Fd, is only applicable when the force is constant and perpendicular to the distance vector. However, in this case, the force is not constant and varies with the angle, so the formula does not apply. Instead, you would need to use the formula for torque due to a varying force, which is Torque (T) = integral of F x dl between limits. In this case, the force is F = BIlsinθ, and the distance vector is dl = r dθ. So, the formula becomes T = integral of BIlsinθ r dθ between limits. Solving this integral, we get T = 2BIrπ. Therefore, the torque due to the varying force is equal to 2BIrπ, and not 4BIπr2 as you have found.