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  • Describe the effects of a magnetic field on a moving charge.
  • Calculate the radius of curvature of the path of a charge that is moving in a magnetic field.

Magnetic force can cause a charged particle to move in a circular or spiral path. Cosmic rays are energetic charged particles in outer space, some of which approach the Earth. They can be forced into spiral paths by the Earth’s magnetic field. Protons in giant accelerators are kept in a circular path by magnetic force. The bubble chamber photograph in [link] shows charged particles moving in such curved paths. The curved paths of charged particles in magnetic fields are the basis of a number of phenomena and can even be used analytically, such as in a mass spectrometer.

A drawing representing trails of bubbles in a bubble chamber.
Trails of bubbles are produced by high-energy charged particles moving through the superheated liquid hydrogen in this artist’s rendition of a bubble chamber. There is a strong magnetic field perpendicular to the page that causes the curved paths of the particles. The radius of the path can be used to find the mass, charge, and energy of the particle.

So does the magnetic force cause circular motion? Magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected, but not the speed. This is typical of uniform circular motion. The simplest case occurs when a charged particle moves perpendicular to a uniform B size 12{B} {} -field, such as shown in [link] . (If this takes place in a vacuum, the magnetic field is the dominant factor determining the motion.) Here, the magnetic force supplies the centripetal force F c = mv 2 / r size 12{F rSub { size 8{c} } = ital "mv" rSup { size 8{2} } /r} {} . Noting that sin θ = 1 size 12{"sin"θ=1} {} , we see that F = qvB size 12{F= ital "qvB"} {} .

Diagram showing an electrical charge moving clockwise in the plane of the page. Velocity vectors are tangent to the circular path. The magnetic field B is oriented into the page. Force vectors show that the force on the charge is toward the center of the charge’s circular path as the charge moves.
A negatively charged particle moves in the plane of the page in a region where the magnetic field is perpendicular into the page (represented by the small circles with x’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, and so velocity changes in direction but not magnitude. Uniform circular motion results.

Because the magnetic force F size 12{F} {} supplies the centripetal force F c size 12{F rSub { size 8{c} } } {} , we have

qvB = mv 2 r . size 12{ ital "qvB"= { { ital "mv" rSup { size 8{2} } } over {r} } "." } {}

Solving for r size 12{r} {} yields

r = mv qB . size 12{r= { { ital "mv"} over { ital "qB"} } "." } {}

Here, r size 12{r} {} is the radius of curvature of the path of a charged particle with mass m size 12{m} {} and charge q size 12{q} {} , moving at a speed v size 12{v} {} perpendicular to a magnetic field of strength B size 12{B} {} . If the velocity is not perpendicular to the magnetic field, then v size 12{v} {} is the component of the velocity perpendicular to the field. The component of the velocity parallel to the field is unaffected, since the magnetic force is zero for motion parallel to the field. This produces a spiral motion rather than a circular one.

Calculating the curvature of the path of an electron moving in a magnetic field: a magnet on a tv screen

A magnet brought near an old-fashioned TV screen such as in [link] (TV sets with cathode ray tubes instead of LCD screens) severely distorts its picture by altering the path of the electrons that make its phosphors glow. (Don’t try this at home, as it will permanently magnetize and ruin the TV.) To illustrate this, calculate the radius of curvature of the path of an electron having a velocity of 6 . 00 × 10 7 m/s size 12{6 "." "00" times "10" rSup { size 8{7} } `"m/s"} {} (corresponding to the accelerating voltage of about 10.0 kV used in some TVs) perpendicular to a magnetic field of strength B = 0 .500 T size 12{B=0 "." "500" T} {} (obtainable with permanent magnets).

A bar magnet with the north pole set against the glass of a computer monitor. The magnetic field lines are shown running from the south pole through the magnet to the north pole. Paths of electrons that are emanating from the computer monitor are shown moving in straight lines until they encounter the magnetic field of the magnet. At that point, they change course and spiral around the magnetic field lines and toward the magnet.
Side view showing what happens when a magnet comes in contact with a computer monitor or TV screen. Electrons moving toward the screen spiral about magnetic field lines, maintaining the component of their velocity parallel to the field lines. This distorts the image on the screen.

Strategy

We can find the radius of curvature r directly from the equation r = m v q B , since all other quantities in it are given or known.

Solution

Using known values for the mass and charge of an electron, along with the given values of v size 12{v} {} and B size 12{B} {} gives us

r = mv qB = 9 . 11 × 10 31 kg 6 . 00 × 10 7 m/s 1 . 60 × 10 19 C 0 . 500 T = 6 . 83 × 10 4 m alignl { stack { size 12{r= { { ital "mv"} over { ital "qB"} } = { { left (9 "." "11" times "10" rSup { size 8{ - "31"} } `"kg" right ) left (6 "." "00" times "10" rSup { size 8{7} } `"m/s" right )} over { left (1 "." "60" times "10" rSup { size 8{ - "19"} } `C right ) left (0 "." "500"`T right )} } } {} #=6 "." "83" times "10" rSup { size 8{ - 4} } `m {} } } {}

or

r = 0 . 683 mm . size 12{r=0 "." "683"" mm"} {}

Discussion

The small radius indicates a large effect. The electrons in the TV picture tube are made to move in very tight circles, greatly altering their paths and distorting the image.

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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