# Current carrying coil - torque on a circular one

1. Apr 20, 2010

### 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 ?

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