# What is the total force on any arbitrary current carrying wire in a uniform magnetic field ?

Sep 14, 2015

$F = B I L \sin \theta$

#### Explanation:

The force on the wire is proportional to:

Current $I$

Length of wire in the field $L$

Sine of angle conductor makes with field $\sin \theta$

The magnetic flux density $B$

Figure 4 shows a wire at an angle $\theta$. The perpendicular vector of the magnetic field that the wire experiences will be equal to $\cos \left(90 - \theta\right)$ = $\sin \theta$.

The force on the wire is given by:

$F = B I L \sin \theta$

From figure 5 you can see that if the wire is perpendicular to the field then $\theta = 90$ and $\sin \theta = 1$.

In this case the force is at a maximum where:

$F = B I L \sin 90$

$= B I L$

In terms of vectors:

$F = B I \stackrel{\rightarrow}{L}$

$\stackrel{\rightarrow}{L}$ is the vector from the start point to the end point. In the case of a closed loop, the wire starts and ends at the same place so:

$\stackrel{\rightarrow}{L} = 0$

So:

$F = B I \times 0 = 0$