# Question 886e9

May 30, 2014

F=BILsinθ

• $B$ is the magnetic field strength (Tesla).
• $I$ is the current in the wire (Amps).
• $L$ is the length of the wire in the magnetic field (metres).
• θ is the angle between the current direction and the magnetic field lines.

The above equation is the general equation for a straight wire in a uniform magnetic field. That will give you the magnitude of the force. In order to find the direction of the force (force is a vector and so has a direction) you will need to use Fleming's Left Hand Rule (FLHR) - see image here https://flic.kr/p/nMRv8W. Magnetic field direction is North to South, and current direction is conventional current direction (positive terminal to negative).

FLHR is based on an angle of 90º between current and magnetic field lines. If θ < 90º then take the component of B-field perpendicular to the current direction and use that as the direction of the field for FLHR.

You are very likely to come across the above equation stated like this:
$F = B I L$
That is true, but it's actually a special case of the equation for when θ=90º#, i.e. $\sin \left(90\right) = 1$ so $B I L \sin \left(90\right) = B I L$.

The most general case, if $\setminus \vec{B} \left(\setminus \vec{r}\right)$ is allowed to vary as a function of position along the length of the wire, and the wire is not straight, is

$\setminus \vec{F} = I \setminus {\int}_{\setminus {\vec{r}}_{0}}^{\setminus {\vec{r}}_{f}} d \setminus \vec{r} \setminus \times \setminus \vec{B} \left(\setminus \vec{r}\right)$

where $I$ is the value of the current in the wire, $\setminus {\vec{r}}_{0}$, and $\setminus {\vec{r}}_{f}$ are the (vector) positions of the beginning and end of the wire, $d \setminus \vec{r}$ is a differential length element along the wire, $\setminus \vec{B} \left(\setminus \vec{r}\right)$ is the magnetic field as a function of position, and $\setminus \times$ denotes the cross (vector) product.

In the case of a closed loop of wire, this integral will be a closed integral with $\setminus {\vec{r}}_{0}$ and $\setminus {\vec{r}}_{f}$ equal to each other, and hence the resulting force will be $\vec{0}$. This does not necessarily mean that the net torque on the wire will be zero though.