svg file showing positive and negative currents flowing in a wire

Electric current, being made up of moving charge, feel a force due to
magnetic fields as well. For example, consider a current flowing to
the right in a magnetic field that points downward. If the current is
made up of positive charges, then according to \(\vec F=q\vec
v\times\vec B\), they will feel a force that's into the page. But
what if the current is made up of electrons, as it usually is in
wires? In that case, the electrons are moving to the left, and \(\vec
v\times\vec B\) points out of the page; but because \(q<0\) we apply
the "electron twist", and the force points into the page. In short,
it doesn't matter what sign the charge carriers are, the direction of
the force is the same: the direction of the current crossed with the
direction of the field.
To write the formula for the force, we need to define a new vector \(\vec I\), which points in the direction of the current with a magnitude equal to \(I\). In terms of this formula, the force on a wire of length \(L\) is

$$\vec F=L\vec I\times\vec B$$

Sometimes you'll see this equation written as
$$\vec F=I\vec L\times\vec B$$
where \(\vec L\) points in the direction of the current and has
magnitude \(\abs{\vec L}=L\). It works pretty much the same way.
Find the force on this current, given this magnetic field.

Consider a rectangular loop of wire, width *W* and height
*H*, which carries a counterclockwise current of *I*.
Find the force on each of the sides of the rectangle, if it is
placed in
(a) a magnetic field \(B\) that points upward
(b) a magnetic field \(B\) that points out of the page

These examples will be important in the next section.

Using the formula \(\vec F=I\vec L\times\vec B\), we see that the
magnetic field and the current run in the same direction along the
right edge, and run in opposite directions on the left edge, so the
cross product on the sides is zero. The magnetic field is
perpendicular to the current along the top and bottom edges, whose
length is \(W\), so the force those edges feel is \(\abs{\vec
F}=IWB\). Using the right-hand rule, we find the following forces:
Notice that the forces all cancel, so the net force is zero. But
something *will* happen to this loop, which we'll discuss in
the next section.