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I = n e A v d .

The magnetic force on any single charge carrier is e v d × B , so the total magnetic force d F on the n A · d l charge carriers in the section of wire is

d F = ( n A · d l ) e v d × B .

We can define dl to be a vector of length dl pointing along v d , which allows us to rewrite this equation as

d F = n e A v d d l × B ,

or

d F = I d l × B .

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.

An illustration of a curving current-carrying wire in a uniform magnetic field. A detail view of a small segment of the wire shows a short, straight piece of current, length d l with current I though it. The velocity v sub d is in the direction of the current. The field B makes an angle theta with the velocity vector.
An infinitesimal section of current-carrying wire in a magnetic field.

To determine the magnetic force F on a wire of arbitrary length and shape, we must integrate [link] over the entire wire. If the wire section happens to be straight and B is uniform, the equation differentials become absolute quantities, giving us

F = I l × B .

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

A wire of length 50 cm and mass 10 g is suspended in a horizontal plane by a pair of flexible leads ( [link] ). The wire is then subjected to a constant magnetic field of magnitude 0.50 T, which is directed as shown. What are the magnitude and direction of the current in the wire needed to remove the tension in the supporting leads?

Figure a: An illustration of a wire suspended in a uniform magnetic field. The magnetic field points into the page. The wire is horizontal and is 50 cm long. A variable voltage source completes the circuit made by the wire and the leads used to suspend it. Figure b: A free body diagram of the wire. The current is to the right. The weight, mg, points down. Tension at either end is up. The magnetic force is up.
(a) A wire suspended in a magnetic field. (b) The free-body diagram for the wire.

Strategy

From the free-body diagram in the figure, the tensions in the supporting leads go to zero when the gravitational and magnetic forces balance each other. Using the RHR-1, we find that the magnetic force points up. We can then determine the current I by equating the two forces.

Solution

Equate the two forces of weight and magnetic force on the wire:

m g = I l B .

Thus,

I = m g l B = ( 0.010 kg ) ( 9.8 m/s 2 ) ( 0.50 m ) ( 0.50 T ) = 0.39 A.

Significance

This large magnetic field creates a significant force on a length of wire to counteract the weight of the wire.

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Calculating magnetic force on a current-carrying wire

A long, rigid wire lying along the y -axis carries a 5.0-A current flowing in the positive y -direction. (a) If a constant magnetic field of magnitude 0.30 T is directed along the positive x -axis, what is the magnetic force per unit length on the wire? (b) If a constant magnetic field of 0.30 T is directed 30 degrees from the + x -axis towards the + y -axis, what is the magnetic force per unit length on the wire?

Strategy

The magnetic force on a current-carrying wire in a magnetic field is given by F = I l × B . For part a, since the current and magnetic field are perpendicular in this problem, we can simplify the formula to give us the magnitude and find the direction through the RHR-1. The angle θ is 90 degrees, which means sin θ = 1 . Also, the length can be divided over to the left-hand side to find the force per unit length. For part b, the current times length is written in unit vector notation, as well as the magnetic field. After the cross product is taken, the directionality is evident by the resulting unit vector.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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