Magnetic Force on a Current-Carrying Conductor

Learning objectives

By the end of this section, you will be able to:

• Determine the direction in which a current-carrying wire experiences a force in an external magnetic field
• Calculate the force on a current-carrying wire in an external magnetic field

Magnetic Fields Produced by Electrical Currents

When discussing historical discoveries in magnetism, we mentioned Oersted’s finding that a wire carrying an electrical current caused a nearby compass to deflect. A connection was established that electrical currents produce magnetic fields. (This connection between electricity and magnetism is discussed in more detail in Sources of Magnetic Fields.)

The compass needle near the wire experiences a force that aligns the needle tangent to a circle around the wire. Therefore, a current-carrying wire produces circular loops of magnetic field. To determine the direction of the magnetic field generated from a wire, we use a second right-hand rule. In RHR-2, your thumb points in the direction of the current while your fingers wrap around the wire, pointing in the direction of the magnetic field produced (CNX_UPhysics_28_04_Wire). If the magnetic field were coming at you or out of the page, we represent this with a dot. If the magnetic field were going into the page, we represent this with an $\times .$ These symbols come from considering a vector arrow: An arrow pointed toward you, from your perspective, would look like a dot or the tip of an arrow. An arrow pointed away from you, from your perspective, would look like a cross or an $\times .$ A composite sketch of the magnetic circles is shown in CNX_UPhysics_28_04_Wire, where the field strength is shown to decrease as you get farther from the wire by loops that are farther separated.

Calculating the Magnetic Force

Electric current is an ordered movement of charge. A current-carrying wire in a magnetic field must therefore experience a force due to the field. To investigate this force, let’s consider the infinitesimal section of wire as shown in CNX_UPhysics_28_04_Infwire. The length and cross-sectional area of the section are dl and A, respectively, so its volume is $V=A·dl.$ The wire is formed from material that contains n charge carriers per unit volume, so the number of charge carriers in the section is $nA·dl.$ If the charge carriers move with drift velocity ${\overrightarrow{v}}_{\text{d}},$ the current I in the wire is (from Current and Resistance)

The magnetic force on any single charge carrier is $e{\overrightarrow{v}}_{\text{d}}\times \overrightarrow{B},$ so the total magnetic force $d\overrightarrow{F}$ on the $nA·dl$ charge carriers in the section of wire is

We can define dl to be a vector of length dl pointing along ${\overrightarrow{v}}_{\text{d}},$ which allows us to rewrite this equation as

or

This is the magnetic force on the section of wire. Note that it is actually the net force exerted by the field on the charge carriers themselves. The direction of this force is given by RHR-1, where you point your fingers in the direction of the current and curl them toward the field. Your thumb then points in the direction of the force.

To determine the magnetic force $\overrightarrow{F}$ on a wire of arbitrary length and shape, we must integrate fs-id1171360233168 over the entire wire. If the wire section happens to be straight and B is uniform, the equation differentials become absolute quantities, giving us

This is the force on a straight, current-carrying wire in a uniform magnetic field.

Balancing the Gravitational and Magnetic Forces on a Current-Carrying Wire

Strategy

Solution

Thus,

Significance

Calculating Magnetic Force on a Current-Carrying Wire

Strategy

Solution

1. We start with the general formula for the magnetic force on a wire. We are looking for the force per unit length, so we divide by the length to bring it to the left-hand side. We also set $\text{sin}\theta =1.$ The solution therefore is $\begin{array}{ccc}F& =& IlB\text{sin}\theta \\ \frac{F}{l}& =& \left(5.0\text{A}\right)\left(\text{0.30 T}\right)\\ \frac{F}{l}& =& 1.5\text{N/m.}\end{array}$ Directionality: Point your fingers in the positive y-direction and curl your fingers in the positive x-direction. Your thumb will point in the $\text{−}\overrightarrow{k}$ direction. Therefore, with directionality, the solution is $\frac{\overrightarrow{F}}{l}=\mathrm{-1.5}\overrightarrow{k}\text{N/m.}$
2. The current times length and the magnetic field are written in unit vector notation. Then, we take the cross product to find the force: $\begin{array}{ccc}\overrightarrow{F}& =& I\overrightarrow{l}\times \overrightarrow{B}=\left(5.0A\right)l\hat{j}\times \left(0.30T\text{cos}\left(30\text{°}\right)\hat{i}+0.30T\text{sin}\left(30\text{°}\right)\hat{j}\right)\\ \overrightarrow{F}\text{/}l& =& \mathrm{-1.30}\hat{k}\text{N/m.}\end{array}$

Significance

Force on a Circular Wire

Strategy

Solution

where $\theta$ is the angle between the magnetic field direction (+y) and the segment of wire. A differential segment is located at the same radius, so using an arc-length formula, we have:

In order to find the force on a segment, we integrate over the upper half of the circle, from 0 to $\pi .$ This results in:

The lower half of the loop is integrated from $\pi$ to zero, giving us:

The net force is the sum of these forces, which is zero.

Significance

Summary

• An electrical current produces a magnetic field around the wire.
• The directionality of the magnetic field produced is determined by the right hand rule-2, where your thumb points in the direction of the current and your fingers wrap around the wire in the direction of the magnetic field.
• The magnetic force on current-carrying conductors is given by $\overrightarrow{F}=I\overrightarrow{l}\times \overrightarrow{B}$ where I is the current and l is the length of a wire in a uniform magnetic field B.

Conceptual Questions

Problems