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Physics 108

Outline - Magnetic Fields
  1. Magnetic Fields and Forces

    1. Forces on charged particles in a magnetic field B
      1. Force F on a charge +q moving with a velocity v in a magnetic field B is F = q(v x B).  F =qvB sin symbol for anglev,B.  The direction of F is given by pointing the fingers of your right hand in the direction of v with the palm of your hand toward B and then rotating the palm of your hand into the direction of B. The thumb of your right hand points in the direction of the force. In Fig. 1 below, with the fingers of your right hand in the direction of v when you rotate your palm toward B, your thumb points out of the page in the direction of F.  A circle with a dot in it, represents a vector out of the page.  A circle with an x in it, represents a vector into the page.  F is always perpendicular to v and B.  F is always perpendicular to the plane of v and B. In Fig. 1, the plane of v and B is the plane of the paper.



      2. A negative charge moving to the right is the same as a positive charge moving to the left. For a negative charge moving to the right the magnetic force would be into the page.



      3. For Fig. 2 above, a positively charged particle of charge +q and mass m travelling with a velocity v to the right enters a magnetic field B into the page. The magnetic force F at the instant shown in the figure is up perpendicular to v and to B. Since there is no component of the force in the direction of the velocity, there is no change in the magnitude of the velocity, but there is a change in its direction. The particle travels in a circle with uniform circular motion.
                     Fnet = ma
         qvB sin 90o = mv2/R
        qvB = mv2/R = 4p2mR/T2 = 4p2mRf2
        since the period T = 2pR/v = 1/f = 1/frequency.


    2. Forces on current-carrying wires in a magnetic field B

      1. Imagine a charge q moving with velocity v in a conductor of length  scripted el.  In time t, the charge moves  scripted el= vt. Since v =  scripted el/t, the magnetic force F = qvB sin symbol for anglev,  B = q( scripted el/t)B sin symbol for anglescripted el bolded, B
        =        (q/t)( scripted el B) sin symbol for anglescripted el bolded,  B = I scripted elB sin symbol for anglescripted el bolded, B. F = I(scripted el boldedx B). The direction of scripted el boldedis the same as the sense of the current I.

      2. Point the fingers of your right hand in the direction of  scripted el boldedwith the palm of your hand toward B and then rotating the palm of your hand into the direction of B. The thumb of your right hand points in the direction of the force F.  In Fig. 3 below, with the fingers of your right hand in the direction of  scripted el boldedwhen you rotate your palm toward B, your thumb points out of the page in the direction of F.



    3. Practice Problems in 108 Problem Set for Magnetic Fields:  2 - 12.



  2. Torque on a Current Carrying Loop in a Magnetic Field


    1. For Fig. 4a above, there will be equal forces up and down on the upper and lower parts of the loop, respectively, but they will not produce any torque about the Y-axis. The force on length cd with current I, scripted el bolded prime =  – scripted el primej and B = Bi is F = I(- scripted el prime j x Bi) = I scripted el prime B k. The torque t = (r x F), where r is drawn from the axis to the point of application of the force F.

    2. It is easier to find the torque by looking at this in the X-Z plane, as shown in Fig. 4b above. The angle between r and F is symbol for angler,F.

      t = rF sin symbol for angler,F
          = scripted el(I scripted el primeB) sin symbol for angler,F = (scripted elscripted el prime)B sin symbol for angler,F
         =      {(area of coil)I]B sin symbol for angler,F.

      If there are N turns,

      t={N(area of coil)I]B sin symbol for angler,F =µB sin symbol for angleµ,B

      where µ= NI(Area) and the direction of µ is found by curling the fingers of your right hand in the direction of the current. Your thumb then points in the direction of µ. Thus,  t = r x F = µx B. For the case shown in Fig. 4b, t is in the negative y or -j direction. In solving problems, the quickest way to find the torque is with t = µ x B.

    3. The magnetic energy of a dipole in a magnetic field B is given by
    4. Practice Problems in 108 Problem Set for Magnetic Fields:   21, 32.



  3. Calculation of Magnetic Fields due to Current Distributions

    1. Magnetic Field due to a very long wire carrying current I

      1. The magnitude of the field B = µoI/2pr, where r is the perpendicular distance from the wire to the point at which you wish to find the magnetic field.  µo = 4px 10-7 N/A2.

      2. Point the thumb of your right hand in the direction of the current. The direction in which your fingers curl is the direction of the magnetic field lines. For a very long wire, the magnetic field lines are concentric circles with the wire at the center of the circles.

        1. Look in the direction of the oncoming current in Fig. 5a below. The magnetic field line is counterclockwise. The magnetic field is tangent to the field line. Above the wire, the field is out of the page. Below the wire, the field is into the page.



        2. It is easier to see this in Fig. 5b below. I have rotated Fig. 5a through 90o and drawn the current out of the page.



      3. The magnetic field due to a very long current-carrying wire is to magnetism what the electric field due to a point charge was to electricity.  Know B = µoI/2pr and how to find the direction of B.

      4. Frequently you will be asked to find the magnetic field due to two or more long current-carrying wires. Always draw the magnetic field lines due to each wire in finding the solution.

        1. In Fig. 6a below wire #1 carries a current I1 = 30 mA and wire #2 carries a current I2 =40 mA. Both currents are out of the page. Find the resultant magnetic field B at point P at the center of the square (Fig. 6a). The square has sides = 20 cm.



        2. The magnetic field due to a long current wire = µoI/2pr.
          µo/2p = 2 x 10-7 N/A2.  For both wires, r = 10 cm = 0.10 m.

          B1 = 2 x 10-7 N/A2 (30 x 10-3 A/0.10 m)
               = 60 x 10-9 N/A-m = 60nT.

          B2 = 2 x 10-7 N/A2 (40 x 10-3 A/0.10 m)
               = 80 x 10-9 N/A-m = 80nT.

          To find the direction of the magnetic fields due to I1 and I2 at point P, draw the magnetic field lines due to each wire such that they pass through P.  If you point the thumb of your right hand in the direction of I1, you find that the magnetic field line due to this current is counterclockwise.  At P (Fig. 6b below), the tangent to the curve (the perpendicular to the radius) is up.  Again from Fig. 6b, you see the magnetic field line due to I2 is counterclockwise and the tangent to the field line at P is to the right. The resultant field at P = B =(602 + 802)1/2 nT = 100 nT.  tan Q = B1/B2 = 3/4. Q = 37o.



        3. Now show that if wire #2 is placed at P' in Fig. 6b above with the current into the page, you get the same result as found in b. above.

      5. Practice Problems in 108 Problem Set for Magnetic Fields:  14-19.

    2. The Biot-Savart Law states 
      is a unit vector drawn from d scripted el bolded to P  and  d scripted el bolded is in the direction of the current I.






      1. Sample Problem

        An otherwise infinite, straight, wire has two concentric loops of radii a and b carrying equal currents in opposite directions as shown in Fig. 8 below. Show that the magnitude of the magnetic field at P, the common center of the loops, is zero if a/b = p/(1 + p).  Try to solve the problem before looking at the answer below. The idea for this problem comes from Kenneth Ford, Classical and Modern Physics (Zerox College Publishing).





      2. Solution to Sample Problem




        1. A to C is 0o and sin 0o = 0.
          There is no contribution from A to C.

        2. D to E is 180o and sin 180o = 0.
          There is no contribution from D to E.

        3. C to D is 90o and sin 90o = 1.  From C to D,
          dB = µoId /4pa2
          and since the length of the arc from C to D = 2pa,
          the total field at P is
          B = µoI(2pa)/4pa2 = µoI/2a.
          Similarly, the field at P due to AE is µoI/2b.
          The field due to the infinite wire is = µoI/2pb.
          The field due to AE and the infinite wire is out of the page and the field due to CD is into the page.

          The field at P is zero if = µoI/2b + µoI/2pb = µoI/2a  or  if
          1/b + 1/pb = (p+ 1)/pb = 1/a  or  a/b = p/(1 + p).

      3. Practice Problems: 108 Problem Set for Magnetism:  28, 29


    3. Ampere's Law

      Ampere's law states that the path integral of the magnetic field around a closed path is equal to µo times the current enclosed by the path.



      Use Ampere's law when B is constant everywhere along the path
      and the  B,d scripted el bolded is 0,  90o, 180o, or 270o.


  4. Application of Ampere's law to

    1. A very long current carrying wire gives B = µoI/2pr

    2. A very long solenoid gives B = µonI,  where I is the current
      and n = the number of turns per unit length.

    3. A toroid of inner radius a and outer radius b gives for

      1. r < a,  B = 0

      2. a < r < b,  B = µoNI/2pr,  where N = the number of turns of the toroid.

    4. Practice Problems in 108 Problem Set for Magnetic Fields: 23-25, 28-30.


  5. Forces between two current-carrying wires

    We wish to find the force of the long wire carrying current  I1 on the long current-carrying wire with current  I2 shown in Fig. 9a below.





    1. The field due to  I1 is Bdue to 1 = µoI1/2p r.  With I1 out of the page, the magnetic field line due to it is counterclockwise.

      The force on 2 = I2 scripted elBdue to 1 sin symbol for anglescripted el, Bdue to 1.

      Since I2 is out of the page and Bdue to 1  is in the page of the paper, the angle is 90o.

      The force per unit length on wire 2 = F on 2/scripted el = I2Bdue to 1
      = I2(µoI1/2p r).  If you point the fingers of your right hand in the direction of  I2 and rotate your fingers into Bdue to 1 your thumb points to the right in the direction of the force on 2.

      Now try to repeat the above finding the field due to 2 at the position of wire #1. Then find the force on 1 due to Bdue to 2, the magnetic field from the current in wire 2. You will find the force on wire #1 is to the left. Newton's third law is still in action.  By experiment, we find that two wires carrying current in the same direction attract each other.



    2. It is more difficult, but not impossible, to visualize the attraction between the two wires if they are drawn as shown in Fig. 9c above.  As you point your finger in the direction of  I1 and you look from above  I1,  you see the magnetic field line is counterclockwise and at the position of the second wire, the field is into the page. Then with  I2 up and Bdue to 1  into the page, the force on 2 is to the right.  Now try it for the force on 1.


    3. Practice Problems in 108 Problem Set for Magnetic Fields: 13, 20, 23, and 27.

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Susan D. Kunk
Phyllis J. Fleming
August 8, 2002
April 30, 2003