12.3 Magnetic Force between Two Parallel Currents

You might expect that two current-carrying wires generate significant forces between them, since ordinary currents produce magnetic fields and these fields exert significant forces on ordinary currents. But you might not expect that the force between wires is used to define the ampere. It might also surprise you to learn that this force has something to do with why large circuit breakers burn up when they attempt to interrupt large currents.

The force between two long, straight, and parallel conductors separated by a distance r can be found by applying what we have developed in the preceding sections. Figure 12.9 shows the wires, their currents, the field created by one wire, and the consequent force the other wire experiences from the created field. Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force $F sub 2$ ). The field due to $I sub 1$ at a distance r is

$B_{1} = \frac{μ_{0} I_{1}}{2 π r}$

(12.9)

Figure A shows two long, straight, and parallel conductors separated by a distance r. The magnetic field produced by one of the conductors is perpendicular to the direction of the flow of the current. Figure b is the top view. It shows that vector F2 is directed from one of the conductors to another. Vector B1 lies at the same plane as the magnetic field and is perpendicular to F2. — **Figure 12.9** (a) The magnetic field produced by a long straight conductor is perpendicular to a parallel conductor, as indicated by right-hand rule (RHR)-2. (b) A view from above of the two wires shown in (a), with one magnetic field line shown for wire 1. RHR-1 shows that the force between the parallel conductors is attractive when the currents are in the same direction. A similar analysis shows that the force is repulsive between currents in opposite directions.

This field is uniform from the wire 1 and perpendicular to it, so the force $F sub 2$ it exerts on a length l of wire 2 is given by $F = I l B \sin θ$ with $sin θ equals 1:$

$F sub 2 equals I sub 2 l B sub 1.$

(12.10)

The forces on the wires are equal in magnitude, so we just write F for the magnitude of $F sub 2.$ (Note that ${\vec{F}}_{1} = - {\vec{F}}_{2} .$ ) Since the wires are very long, it is convenient to think in terms of F/l, the force per unit length. Substituting the expression for $B sub 1$ into Equation 12.10 and rearranging terms gives

$\frac{F}{l} = \frac{μ_{0} I_{1} I_{2}}{2 π r} .$

The ratio F/l is the force per unit length between two parallel currents $I sub 1$ and $I sub 2$ separated by a distance r. The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.

Function graph showing the direction field of the system x′ = -y/((x+1)^2+y^2) - I2*y/((x-1)^2+y^2), y′ = (x+1)/((x+1)^2+y^2) + I2*(x-1)/((x-1)^2+y^2), drawn from 308 sample arrows, and 10 trajectories through (-1, 0.22), (-1, 0.5), (-1, 0.95), (-1, 1.6), (-1, -0.22), (-1, -0.5), (-1, -0.95), (1, 0.3), (1, -0.3) and (0, 2.1). Adjustable parameter: Current in the right-hand wire, in units of the left-hand current (I2) = 1. Viewing window: x from -3.8 to 3.8, y from -2.35 to 2.35. — The combined field of two long parallel wires, each contributing B = μ₀I/2πr wrapped around itself: one carries current I₁ out of the page at x = −1, the other sits at x = +1 and its current is the slider, measured in units of I₁. Distances are in units of the half-separation. The arrows are the field itself and the ten curves are streamlines started around the left wire, so the closed loops you see are computed magnetic field lines. The default +1 is the arrangement in the example Calculating Forces on Wires — both wires carrying the same current the same way, which for its two 5.0 mA wires 5.0 cm apart gives F/l = 1 × 10⁻¹⁰ N/m of attraction. Look at what the picture says about that force. With both currents alike, the two contributions cancel at the midpoint and each wire sits in the other's field pointing so that IL × B pulls it inward. Slide the current to −1 and the midpoint field doubles instead of cancelling, the loops stop enclosing both wires and start threading between them, and the force reverses — attraction for like currents, repulsion for opposite ones, exactly as the sentence above says. At 0 only the left wire's circles remain.

This force is responsible for the pinch effect in electric arcs and other plasmas. The force exists whether the currents are in wires or not. It is only apparent if the overall charge density is zero; otherwise, the Coulomb repulsion overwhelms the magnetic attraction. In an electric arc, where charges are moving parallel to one another, an attractive force squeezes currents into a smaller tube. In large circuit breakers, such as those used in neighborhood power distribution systems, the pinch effect can concentrate an arc between plates of a switch trying to break a large current, burn holes, and even ignite the equipment. Another example of the pinch effect is found in the solar plasma, where jets of ionized material, such as solar flares, are shaped by magnetic forces.

The definition of the ampere is based on the force between current-carrying wires. Note that for long, parallel wires separated by 1 meter with each carrying 1 ampere, the force per meter is

$\frac{F}{l} = \frac{(4 π \times 10^{−7} T \cdot m/A) {(1 A)}^{2}}{(2 π) (1 m)} = 2 \times 10^{−7} N/m .$

Since $μ sub 0$ is exactly $4 π \times 10^{−7} T \cdot m/A$ by definition, and because $1 T = 1 N/ (A \cdot m),$ the force per meter is exactly $2 \times 10^{−7} N/m .$ This is the basis of the definition of the ampere.

Infinite-length wires are impractical, so in practice, a current balance is constructed with coils of wire separated by a few centimeters. Force is measured to determine current. This also provides us with a method for measuring the coulomb. We measure the charge that flows for a current of one ampere in one second. That is, $1 C = 1 A \cdot s .$ For both the ampere and the coulomb, the method of measuring force between conductors is the most accurate in practice.

Example 1 Calculating Forces on Wires

Two wires, both carrying current out of the page, have a current of magnitude 5.0 mA. The first wire is located at (0.0 cm, 3.0 cm) while the other wire is located at (4.0 cm, 0.0 cm) as shown in Figure 12.10. What is the magnetic force per unit length of the first wire on the second and the second wire on the first?

Figure 12.10 Two current-carrying wires at given locations with currents out of the page.

Strategy Each wire produces a magnetic field felt by the other wire. The distance along the hypotenuse of the triangle between the wires is the radial distance used in the calculation to determine the force per unit length. Since both wires have currents flowing in the same direction, the direction of the force is toward each other.

Solution The distance between the wires results from finding the hypotenuse of a triangle:

$r = \sqrt{{(3.0 cm)}^{2} + {(4.0 cm)}^{2}} = 5.0 cm .$

The force per unit length can then be calculated using the known currents in the wires:

$\frac{F}{l} = \frac{(4 π \times 10^{−7} T \cdot m/A) {(5 \times 10^{−3} A)}^{2}}{(2 π) (5 \times 10^{−2} m)} = 1 \times 10^{−10} N/m .$

The force from the first wire pulls the second wire. The angle between the radius and the x-axis is

$θ = \tan^{−1} (\frac{3 cm}{4 cm}) = 36.9 ° .$

The unit vector for this is calculated by

$- \cos ({36.9}^{\circ}) \hat{i} + \sin ({36.9}^{\circ}) \hat{j} = –0.8 \hat{i} + 0.6 \hat{j} .$

Therefore, the force per unit length from wire one on wire 2 is

$\frac{\vec{F}}{l} = (1 \times 10^{−10} N/m) \times (–0.8 \hat{i} + 0.6 \hat{j}) = (–8 \times 10^{−11} \hat{i} + 6 \times 10^{−11} \hat{j}) N/m .$

The force per unit length from wire 2 on wire 1 is the negative of the previous answer:

$\frac{\vec{F}}{l} = (8 \times 10^{−11} \hat{i} - 6 \times 10^{−11} \hat{j}) N/m .$

Significance These wires produced magnetic fields of equal magnitude but opposite directions at each other’s locations. Whether the fields are identical or not, the forces that the wires exert on each other are always equal in magnitude and opposite in direction (Newton’s third law).

Summary

The force between two parallel currents $I sub 1$ and $I sub 2,$ separated by a distance r, has a magnitude per unit length given by $\frac{F}{l} = \frac{μ_{0} I_{1} I_{2}}{2 π r} .$
The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

Conceptual Questions

Compare and contrast the electric field of an infinite line of charge and the magnetic field of an infinite line of current.

Is $\vec{B}$ constant in magnitude for points that lie on a magnetic field line?

A magnetic field line gives the direction of the magnetic field at any point in space. The density of magnetic field lines indicates the strength of the magnetic field.

Problems

Two long, straight wires are parallel and 25 cm apart. (a) If each wire carries a current of 50 A in the same direction, what is the magnetic force per meter exerted on each wire? (b) Does the force pull the wires together or push them apart? (c) What happens if the currents flow in opposite directions?

Two long, straight wires are parallel and 10 cm apart. One carries a current of 2.0 A, the other a current of 5.0 A. (a) If the two currents flow in opposite directions, what is the magnitude and direction of the force per unit length of one wire on the other? (b) What is the magnitude and direction of the force per unit length if the currents flow in the same direction?

a. $F / l = 2 \times 10^{−5} N/m$ away from the other wire; b. $F / l = 2 \times 10^{−5} N/m$ toward the other wire

Two long, parallel wires are hung by cords of length 5.0 cm, as shown in the accompanying figure. Each wire has a mass per unit length of 30 g/m, and they carry the same current in opposite directions. What is the current if the cords hang at $6.0 °$ with respect to the vertical?

Figure shows two parallel wires with current flowing in opposite directions that are hung by cords suspended from hooks.

A circuit with current I has two long parallel wire sections that carry current in opposite directions. Find magnetic field at a point P near these wires that is a distance a from one wire and b from the other wire as shown in the figure.

Figure shows two current carrying wires. One carries current out of the page; another carries current into the page. Wires form vertices of a right triangle. Point P is the third vertex and is located at a distance b from one wire and distance a from another wire. Distance b is a leg; distance a is a hypotenuse.

$B = \frac{μ_{o} I}{2 π a^{2} b} ((a_{2} - b_{2}) \hat{i} + b \sqrt{(a^{2} - b^{2})} \hat{j})$

The infinite, straight wire shown in the accompanying figure carries a current $I sub 1.$ The rectangular loop, whose long sides are parallel to the wire, carries a current $I sub 2.$ What are the magnitude and direction of the force on the rectangular loop due to the magnetic field of the wire?

Figure shows a wire that carries the current I1 and a rectangular loop with long sides that are parallel to the wire and carry a current I2. Distance between the wire and the loop is b. Length of the side of the long side of the loop is a, distance of the short side of the loop is b.