# Force on wire by earth's mag. field

1. Aug 25, 2013

### eroxore

The task is to calculate the force acting on a 50 meter long current-carrying wire due to earth's magnetic field. The angle of inclination is assumed to be 70 degrees. Here's a schematic diagram:

The exaggeratedly drawn wire represents a segment of a wire on a power grid. In order to find the force, we need to find the component of earth's magnetic field perpendicular to the direction of the current in the wire. I began therefore by calculating the vertical component of earth's magnetic field, yielding $B_{vert.} = 50\sin 70^\circ$. Let's now redraw the picture with only the vertical component:

We see now that the perpendicular component of can be calculated as the sin(70) of the vertical component. Thus $B_{\perp} = 50\sin 70^\circ \cdot \sin 70^\circ$. We can finally use $F = IlB$ to calculate the force and we end up with approximately 2.2 N.

This is however wrong and the actual answer is in fact 1.9 N. The "accepted" approach is to use sin(50) of earth's magnetic field. I fully understand this, but what I want to unveil is why my approach failed. What exactly was wrong in my arguments? The math or the physics?

Help very much appreciated!

2. Aug 25, 2013

### Electric Red

Hello, well as you say, you understand what to do.
You indeed need to use $F=BIL$, since $B_{effective}$ is $Bsin(70-20), B_{effective}$ is $50μTsin(50)$
Putting in the correct values give the correct answer of $1.9 N$

Your approach seems a bit odd, you use two times sin(70), I can't figure out which component you want to calculate.

3. Aug 25, 2013

### eroxore

Nevermind, I figured out myself why my approach was failing. I forgot to consider the perpendicular component of the horizontal component of earth's magnetic field vector.