Magnetic Force on a Current-Carrying Wire
Current flows in a wire specifically in conductors. When there is current traveling, a magnetic field will be produced in the surroundings which exerts force to other charges moving or present in the field. Since these charged particles cannot escape out of the wire, it then follows that the wire also experiences magnetic force. The magnetic force exerted on each particle will be transmitted to the wire as the particles collide with the atoms that made up the wire. Since magnetic field exerts force on every particle, hence, we could then say that the resultant magnetic force experienced by the wire is the vector sum of the magnetic forces experienced by each particle in the wire.
The figures below show the behavior of a wire when suspended vertically between the poles of a magnet.
Consider a straight wire segment of length L and cross-sectional area A carrying a current I moving in uniform magnetic field \(\vec B\). The total number of charges in the segment is the product of the number of moving charges per unit volume, n and the volume of the wire, AL. Thus, to solve for the magnetic force in a current-carrying wire, we multiply the equation for magnetic force \(\vec F = q\vec v \times \vec B\) with the total number of charges, nAL. It can be written mathematically as
\(\vec F = (q\vec v \times \vec B)nAL\) (1)
Note that the current in a wire is given by the equation \(I=nqv_dA\). We can rewrite equation (1) as
\(\vec F = I\vec L \times \vec B\) (2)
where \(\vec L\) is a vector whose magnitude is equal to L and points in the direction of the current I. Equation (2) also denotes that the force is always perpendicular to the conductor and the magnetic field. The direction is still determined using the right-hand rule for a moving positively-charged particle. This equation also applies only on straight wires or conductors.
Now, we consider a wire of different shape (not straight) but with uniform cross section in a magnetic field. To solve for the magnetic force in a wire, we consider dividing the wire into different small segments with vector length \(d\vec l\). The magnetic force in each segment of the wire can then be solved using the equation
\(d\vec F = Id\vec l \times \vec B\) (3)
The figure below shows a wire segment of arbitrary shape carrying current I in a magnetic field \(\vec B\).
We will then integrate equation (3) over the length of the wire to solve for the total force acting on the wire as shown in the above figure.
\(\vec F = I \int \limits_a^b d\vec l \times \vec B\) (4)
a and b indicates the endpoints of the wire.