Magnetic Force on a Wire Carrying a Current

Straight wire in a perpendicular field

Diagram (Fig. 10.8): a straight wire of length $L$ lies in the plane of the paper carrying a current $I$; the applied magnetic field $\vec{B}$ is drawn as dots, coming out of the plane of the paper, so it is perpendicular to the wire. The conduction electrons drift with velocity $\vec{v}_d$ opposite to the conventional current. The resulting magnetic force $\vec{F}_m$ on the wire is marked perpendicular to both, in the plane of the paper. A plane pp cuts across the wire.

Let $I$ be the current and $v_d$ the drift velocity of the conduction electrons. The charge crossing the plane pp in time $t$ is $q = It$, and the time for the electrons to traverse the length $L$ is $t = L/v_d$, so

$$q = \dfrac{IL}{v_d}$$

The magnetic force on this charge, from $\vec{F}_m = q(\vec{v}\times\vec{B})$, has magnitude

$$F_m = q\,v_d B \sin 90^\circ = \left(\dfrac{IL}{v_d}\right) v_d B = ILB$$

$$\therefore \ \vec{F}_m = ILB\,\hat{n}$$

where $\hat{n}$ is the unit vector perpendicular to both $\vec{B}$ and $\vec{v}_d$, in the direction of $\vec{F}_m$. This is the force on a straight length $L$ of the wire.

Field not perpendicular to the wire

If $\vec{B}$ makes an angle $\theta$ with the wire, the result generalises to

$$\vec{F}_m = I\,\vec{L}\times\vec{B}, \qquad F_m = ILB\sin\theta$$

where $\vec{L}$ is the length vector, of magnitude $L$, directed along the portion of the wire in the sense of the current.

Arbitrarily shaped wire

Diagram (Fig. 10.9): a wire of arbitrary curved shape carrying current $I$, with a short element $d\vec{l}$ marked on it, sitting in a field $\vec{B}$ coming out of the plane of the paper; the elementary force $d\vec{F}_m$ is drawn perpendicular to $d\vec{l}$.

Consider a segment of infinitesimal length $d\vec{l}$. Applying the straight-wire result to it,

$$d\vec{F}_m = I\,d\vec{l}\times\vec{B}$$

The force on the whole wire is the sum of these:

$$\vec{F}_m = \int d\vec{F}_m = \int I\,d\vec{l}\times\vec{B}$$

If $\vec{B}$ is uniform over the whole wire it can be taken outside the integral:

$$\vec{F}_m = I\left(\int d\vec{l}\right)\times\vec{B}$$

so that only the vector sum of the length elements — that is, the straight line joining the two ends of the wire — determines the force.

Standard results

• $q = \dfrac{IL}{v_d}$ (charge crossing a section in the transit time)
• $F_m = ILB$ for a straight wire perpendicular to a uniform field
• $\vec{F}_m = I\vec{L}\times\vec{B}$, magnitude $ILB\sin\theta$, in general
• $d\vec{F}_m = I\,d\vec{l}\times\vec{B}$ for an element of any wire
• $\vec{F}_m = I\left(\int d\vec{l}\right)\times\vec{B}$ for uniform $\vec{B}$
• Levitation condition (textbook Example 10.3): a loop of arm length $a$ carrying current $I$ with its horizontal arm in the field is held up when $IBa = mg$, so $I = \dfrac{mg}{Ba}$. The same condition levitates a straight wire of mass $m$ and length $L$: $BIL = mg$.