Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
College Physics for AP® Courses 2e

22.7 Magnetic Force on a Current-Carrying Conductor

College Physics for AP® Courses 2e22.7 Magnetic Force on a Current-Carrying Conductor

Learning Objectives

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

  • Describe the effects of a magnetic force on a current-carrying conductor.
  • Calculate the magnetic force on a current-carrying conductor.

Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself.

A diagram showing a circuit with current I running through it. One section of the wire passes between the north and south poles of a magnet with a diameter l. Magnetic field B is oriented toward the right, from the north to the south pole of the magnet, across the wire. The current runs out of the page. The force on the wire is directed up. An illustration of the right hand rule 1 shows the thumb pointing out of the page in the direction of the current, the fingers pointing right in the direction of B, and the F vector pointing up and away from the palm.
Figure 22.29 The magnetic field exerts a force on a current-carrying wire in a direction given by the right hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire, since typical currents consist of very large numbers of moving charges.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vdvd is given by F=qvdBsinθF=qvdBsinθ. Taking BB to be uniform over a length of wire ll and zero elsewhere, the total magnetic force on the wire is then F=(qvdBsinθ)(N)F=(qvdBsinθ)(N), where NN is the number of charge carriers in the section of wire of length ll. Now, N=nVN=nV, where nn is the number of charge carriers per unit volume and VV is the volume of wire in the field. Noting that V=AlV=Al, where AA is the cross-sectional area of the wire, then the force on the wire is F=(qvdBsinθ)(nAl)F=(qvdBsinθ)(nAl). Gathering terms,

F=(nqAvd)lBsinθ.F=(nqAvd)lBsinθ.
22.15

Because nqAvd=InqAvd=I (see Current),

F = IlB sin θ F = IlB sin θ
22.16

is the equation for magnetic force on a length ll of wire carrying a current II in a uniform magnetic field BB, as shown in Figure 22.30. If we divide both sides of this expression by ll, we find that the magnetic force per unit length of wire in a uniform field is Fl=IBsinθFl=IBsinθ. The direction of this force is given by RHR-1, with the thumb in the direction of the current II. Then, with the fingers in the direction of BB, a perpendicular to the palm points in the direction of FF, as in Figure 22.30.

Illustration of the right hand rule 1 showing the thumb pointing right in the direction of current I, the fingers pointing into the page with magnetic field B, and the force directed up, away from the palm.
Figure 22.30 The force on a current-carrying wire in a magnetic field is F=IlBsinθF=IlBsinθ. Its direction is given by RHR-1.

Example 22.4

Calculating Magnetic Force on a Current-Carrying Wire: A Strong Magnetic Field

Calculate the force on the wire shown in Figure 22.29, given B=1.50 TB=1.50 T, l=5.00 cml=5.00 cm, and I=20.0AI=20.0A.

Strategy

The force can be found with the given information by using F=IlBsinθF=IlBsinθ and noting that the angle θθ between II and BB is 90º90º, so that sinθ=1sinθ=1.

Solution

Entering the given values into F=IlBsinθF=IlBsinθ yields

F=IlBsinθ=20.0 A0.0500 m1.50 T1 .F=IlBsinθ=20.0 A0.0500 m1.50 T1 .
22.17

The units for tesla are 1 T=NAm1 T=NAm; thus,

F=1.50 N.F=1.50 N.
22.18

Discussion

This large magnetic field creates a significant force on a small length of wire.

Magnetic force on current-carrying conductors is used to convert electric energy to work. (Motors are a prime example—they employ loops of wire and are considered in the next section.) Magnetohydrodynamics (MHD) is the technical name given to a clever application where magnetic force pumps fluids without moving mechanical parts. (See Figure 22.31.)

Diagram showing a cylinder of fluid of diameter l placed between the north and south poles of a magnet. The north pole is to the left. The south pole is to the right. The cylinder is oriented out of the page. The magnetic field is oriented toward the right, from the north to the south pole, and across the cylinder of fluid. A current-carrying wire runs through the fluid cylinder with current I oriented downward, perpendicular to the cylinder. Negative charges within the fluid have a velocity vector pointing up. Positive charges within the fluid have a velocity vector pointing downward. The force on the fluid is out of the page. An illustration of the right hand rule 1 shows the thumb pointing downward with the current, the fingers pointing to the right with B, and force F oriented out of the page, away from the palm.
Figure 22.31 Magnetohydrodynamics. The magnetic force on the current passed through this fluid can be used as a nonmechanical pump.

A strong magnetic field is applied across a tube and a current is passed through the fluid at right angles to the field, resulting in a force on the fluid parallel to the tube axis as shown. The absence of moving parts makes this attractive for moving a hot, chemically active substance, such as the liquid sodium employed in some nuclear reactors. Experimental artificial hearts are testing with this technique for pumping blood, perhaps circumventing the adverse effects of mechanical pumps. (Cell membranes, however, are affected by the large fields needed in MHD, delaying its practical application in humans.) MHD propulsion for nuclear submarines has been proposed, because it could be considerably quieter than conventional propeller drives. The deterrent value of nuclear submarines is based on their ability to hide and survive a first or second nuclear strike. As we slowly disassemble our nuclear weapons arsenals, the submarine branch will be the last to be decommissioned because of this ability (See Figure 22.32.) Existing MHD drives are heavy and inefficient—much development work is needed.

Diagram showing a zoom in to a magnetohydrodynamic propulsion system on a nuclear submarine. Liquid moves through the thruster duct, which is oriented out of the page. Magnetic fields emanate from the coils and pass through a duct. The magnetic flux is oriented up, perpendicular to the duct. Each duct is wrapped in saddle-shaped superconducting coils. An electric current runs to the right, through the liquid and perpendicular to the velocity of the liquid. The electric current flows between a pair of electrodes inside each thruster duct. A repulsive interaction between the magnetic field and electric current drives water through the duct. An illustration of the right hand rule shows the thumb pointing to the right with the electric current. The fingers point up with the magnetic field. The force on the liquid is oriented out of the page, away from the palm.
Figure 22.32 An MHD propulsion system in a nuclear submarine could produce significantly less turbulence than propellers and allow it to run more silently. The development of a silent drive submarine was dramatized in the book and the film The Hunt for Red October.
Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/college-physics-ap-courses-2e/pages/1-connection-for-ap-r-courses
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/college-physics-ap-courses-2e/pages/1-connection-for-ap-r-courses
Citation information

© Jul 9, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.