Why does F=BIL sin θ find the component perpendicular to B?

[adamg]

The force on a current carrying conducter is F=BIL sin@ . One book tells me that you do this because you need to find the component of I at 90 degrees to B. Yet another says you need to find the component of L at 90 degrees to B. Obviously you don't work out BxIsin@xLsin@, so you don't work out both components.

So I was wondering, are you really just working out the component of the quantity IxL that is at right angles to B?

 

[Doc Al]

Don't think of "I" and "L" as separate vectors. The current (I) follows the wire (L).

Think of the force as F = I L X B = ILBsinθ; where θ is the angle between L and B; I is a scalar.

 

[adamg]

Yeah. I guess if you work out the vertical component of the length, that is the vertical distance through which the current has travelled, i.e. the distance through which it is at 90 degrees to B, and the horizontal distance can be forgotten as it is parallel to B. That makes sense.