Chapter 28 Magnetic Fields

Document Sample
Chapter 28 Magnetic Fields Powered By Docstoc
					                      Chapter 28
                     Magnetic Fields

In this chapter we will cover the following topics:

                        r
Magnetic field vector, B
                                     r
Magnetic force on a moving charge, FB
Magnetic field lines
M      ti fi ld li
Motion of a moving charge particle in a uniform magnetic field
Magnetic force on a current-carrying wire
Magnetic torque on a wire loop
                                           r
Magnetic dipole, magnetic dipole moment μ
Hall effect
Cyclotron particle accelerator
                                                       (28-1)
         What Produces a Magnetic Field
         One can generate a magnetic field using one of the
         following th d
         f ll i methods:


         Pass a current through a wire and thus form what is known
         as an "electromagnet."


                permanent magnet.
         Use a "permanent" magnet


         Empirically we know that both types of magnets attract
         small pieces of iron. Also, if suspended so that they can
         rotate freely they align themselves along the north-south
                                    y              g
         direction. We can thus say that these magnets create in
                                                     r
         the surrounding space a "magnetic field" B, which
                                                       r
         manifests itself by exerting a magnetic force FB .
         We will use the magnetic force to define precisely
                                   r
         the magnetic field vector B.
(28-2)
                                                      r     r r
              r               FB = q vB sin φ         FB = qv × B
Definition of B
                                                             r
The magnetic field vector is defined in terms of the force FB it
                                                   r
exerts on a charge q, which moves with velocity v . We inject
                                                         r
the charge q in a region where we wish to determine B at
random directions, trying to scan all the possible directions.
                                            r
There is one direction for which the force FB on q is zero. This
                           r                            r
di ti is parallel with B . For all other di ti
direction i     ll l ith      F                            i   t
                                   ll th directions FB is not zero,
and its magnitude FB = q v sin φ where φ is the angle between
r      r              r
v and B. In addition, FB is perpendicular to the plane defined
                    ,       p p                  p
   r      r
by v and B. The magnetic force vector is given by the equation
 r     r r
FB = qv × B.
SI unit of B : The defining equation is FB = q v sin φ .
If we shoot a particle with charge q = 1 C at right
                     r
angles (φ = 90°) to B with speed v = 1m/s and the
magnetic force FB = 1 N, then B = 1 tesla.
                                                           (28-3)
The Vector Product of Two Vectors
                   r r r                    r     r
The vector product c = a × b of the vectors a and b
            r
is a vector c .
                  r
The magnitude of c is given by the equation
c = ab sin φ .
                   r
The direction of c is perpendicular to the plane P defined
                r     r
by the vectors a and b .
                         r
The sense of the vector c is given by the right - hand rule:
                      r       r
a. Place the vectors a and b tail to tail.
            r
b. Rotate a in the plane P along the shortest angle
                          r
so that it coincides with b .
c Rotate the fingers of the right hand in the same direction
c.                                                 direction.
                                                    r
d. The thumb of the right hand gives the sense of c .
The vector product of two vectors is also known as
the "cross" product.
                                                     (28-4)
                   r r r
The Vector Product c = a × b in Terms of Vector Components

r                         r
a = a xi       j      ˆ           ˆ      j      ˆ r
       ˆ + a y ˆ + az k , b = b x i + by ˆ + bz k , c = c x i + c y ˆ + cz k
                                                            ˆ       j      ˆ
                                       r
The vector components of vector c are given by the equations
cx = a y bz − az by , c y = az bx − ax bz , cz = ax by − a y bx


Note : Those familiar with the use of determinants can use the expression
         $
         i   j $
            $ k
 r r
a × b = ax a y az
          bx   by       bz


Note : The order of the two vectors in the cross product is important:
r r       r r
          (
b × a = − a ×b .    )
                                                                               (28-5)
Magnetic Field Lines : In analogy with the electric field lines we
introduce the concept of magnetic field lines, which help visualize
                          r
the        ti field   t      ith t i            ti
th magnetic fi ld vector B without using equations.
                                                     r
In the relation between the magnetic field lines and B :
                                                r
          yp               g                           g            g
1. At any point P the magnetic field vector B is tangent to the magnetic field lines.
                                              r
                                              BP
                                         g
                                       magnetic field line
                             P
                                              r
2. The magnitude of the magnetic field vector B is proportional
to the density of the magnetic field lines.
                                              r           r
        BP > BQ                               BP          BQ
                                                      Q
                            P                                   magnetic field lines

    (28-6)
         Magnetic Field Lines of a Permanent Magnet
         The magnetic field lines of a permanent magnet are shown
         in the figure. The lines pass through the body of the magnet
         and form closed loops. This is in contrast to the electric field
         lines that originate and terminate on electric charges.
                       g                                    g
         The closed magnetic field lines enter one point of the magnet
         and exit at the other end. The end of the magnet from which
         th lines emerge is known as the north pole of the magnet.
         the li            i k         th     th l f th          t
         The other end where the lines enter is called the south pole
         of the magnet. The two poles of the magnet cannot be
         separated. Together they form what is known as a
          "magnetic dipole."




(28-7)
                       r
                       FE
    Cathode
                                                                       r     r
                                                                       FE = qE
                 Anode
                         r                                            r     r r
                         FB                                           FB = qv × B

                                                                   figure
Discovery of the Electron : A cathode ray tube is shown in the figure. Electrons are
emitted from a hot filament known as the "cathode." They are accelerated by a
voltage V applied between the cathode and a second electrode known as the "anode."
The electrons pass through a hole in the anode and they form a narrow beam. They
hit the fluorescent coating of the right wall of the cathode ray tube where they produce
          light J J
a spot of light. J.J. Thomson in 1897 used a version of this tube to investigate the nature
of the particle beam that caused the fluorescent spot. He applied constant electric and
magnetic fields in the tube region to the right of the anode. With the fields oriented as
                                       r                          r
shown in the figure the electric force FE and the magnetic force FB have opposite
directions. By adjusting B and E , Thomson was able to have a zero net force.
                                                                                 (28-8)
                r
                v     r
                 r
                     .B         Motion of a Charged Particle in a Uniform Magnetic Field
                     electron   (also known as cyclotron motion)
                F
                                A particle of mass m and charge q, when injected with a speed
                                       i l f             d h          h i j       d ih      d
        C   .                                                                 r
                                v at right angles to a uniform magnetic field B, follows a
                r
                                                             p               p
                                circular orbit with uniform speed. The centripetal force
                                required for such motion is provided by the magnetic force
                                         r     r r
r=
     mv    qB                           FB = qv × B
        ω=
     qB    m

The circular orbit of radius r for an electron is shown in the figure. The magnetic force
                  v2     mv                     2π r 2π mv 2π m
FB = q vB = ma = m → r =    . The period is T =     =      =    .
                  r      qB                      v    q Bv   qB
                                    1    qB                                        qB
The corresponding frequency is f =    =     . The angular frequency is ω = 2π f =     .
                                   T 2π m                                          m
Note 1 : The cyclotron period does not depend on the speed v. All particles of the
same mass complete their circular orbit during the same time T regardless of speed.
             p                               g                   g            p
Note 2 : Fast particles move on larger-radius circular orbits, while slower particles move
on smaller-radius orbits. All orbits have the same period T .                (28-9)
      mv⊥             2π m                                         Helical Paths
 r=              T=
      qB               qB                                          We now consider the motion of
                                                                            in
                                                                   a charge i a uniform magnetic
                                                                      h            if             i
                                                                         r
                                                                   field B when its initial velocity
                                                                   r                          r
                                                                   v forms an angle φ with B.
                                                                                    r
                                                                   We decompose v into two
                                                                   components.

                                            r                                          r
One component (v || )        is parallel to B and the other ( v⊥ ) is perpendicular to B (see fig. a ):
v || = v cos φ    v⊥ = v sin φThe particle executes two independent motions.
                                                             r
One, the cyclotron motion, is in the plane perpendicular to B that we have
                                                mv                          2π m
analyzed on the previous page. Its radius is r = ⊥ . Its period is T =           .
                                                 qB                          qB
                                              r
The second motion is along the direction of B and it is linear motion with constant
                                                                              fig. ).
speed v ||. The combination of the two motions results in a helical path (see fig b)
      v
                                                       2π mv cos φ
The pitch p of the helix is given by p = Tv || =
                                          v                        .                    (28-10)
                                                           qB
     Magnetic Force on a Current - Carrying Wire
     Consider a wire of length L that carries a current i as shown in
     the figure. A uniform magnetic field B is present in the vicinity
r                                                             r
FB   of the wire. Experimentally it was found that a force FB is
                 r                       r
          t d by     the i       d that     i         di l
     exerted b B on th wire, and th t FB is perpendicular
     to the wire. The magnetic force on the wire is the vector sum
                                            r
     of all the magnetic forces exerted by B on the electrons that
                   g                     y
     constitute i. The total charge q that flows through the wire
     in time t is given by
                L
     q = it = i    . Here vd is the drift velocity of the electrons
                vd
     in the wire.
                                                    L
     The magnetic force is FB = qvd B sin 90° = i      vd B = iLB.
                                                    vd
      FB = iLB

                                                           (28-11)
                           Magnetic Force on a Straight Wire in a Uniform
                           Magnetic Field
                           If we assume the more general case for which the
                                          r
                           magnetic field B forms an angle φ with the wire
                           the magnetic force equation can be written in vector
                             e ag e c o ce equa o ca          w e        vec o
                                   r     r r           r
    r     r r              form as FB = iL × B. Here L is a vector whose
    FB = iL × B
                           magnitude is equal to the wire length L and
                           has a direction that coincides with that of the current.
                  r        The magnetic force magnitude is FB = iLB sin φ .
                  B           g                                y    p
                           Magnetic Force on a Wire of Arbitrary Shape
           r          φ
i
          dF   .       r
                      dL
                           Placed in a Nonuniform Magnetic Field
                           In this case we divide the wire into elements of
 r      r r                                                        straight
                           length dL, which can be considered as straight.
dFB = idL × B              The magnetic force on each element is
                            r       r r
r         r r              dFB = idL × B. The net magnetic force on the
FB = i ∫ dL × B                                          r         r r
                           wire is given by the integral FB = i ∫ dL × B.

                                                                        (28-12)
                                    τ net = iAB sin θ             Side view
                  Top view

                                        C net = 0
                                         F

                                                              C




Magnetic Torque on a Current Loop
Consider the rectangular loop in fig. a with sides of lengths a and b and that carries
                                                                       ˆ
a current i. The loop is placed in a magnetic field so that the normal n to the loop
                        r
forms an angle θ with B. The magnitude of the magnetic force on sides 1 and 3 is
F1 = F3 = iaB sin 90° = iaB. The magnetic force on sides 2 and 4 is
F2 = F4 = ibB sin(90 − θ ) = ibB cos θ . These forces cancel in pairs and thus Fnett = 0.
The torque about the loop center C of F2 and F4 is zero because both forces pass
through point C. The moment arm for F1 and F3 is equal to (b / 2) sin θ . The two
torques tend to rotate the loop in the same (clockwise) direction and thus add up.
The net torque τ = τ 1 +τ 3 =(iabB / 2) sin θ + (iabB / 2) sin θ = iabB sin θ = iAB sin θ .
                                                                                    (28-13)
                                                                              r    r
                                                                                   r
                           Magnetic Dipole Moment                            τ = μ×B
                           The torque of a coil that has N loops exerted           r r
                                                                              U = −μ ⋅ B
                           by a uniform magnetic field B and carries a
                           current i is given by the equation τ = NiAB.
                                                     r
                           We define a new vector μ associated with the coil,
                           which is known as the magnetic dipole moment of
 U = μB       U = −μ B     the coil.
The magnitude of the magnetic dipole moment is μ = NiA.
    magnit de
Its direction is perpendicular to the plane of the coil.
                r
The sense of μ is defined by the right-hand rule. We curl the fingers of the right hand
in the direction of the current. The thumb gives us the sense. The torque can be
                                                                  r       r
expressed in the form τ = μ B sin θ where θ is the angle between μ and B.
                  r r r
In vector form: τ = μ × B.
                                                         r r
The potential energy of the coil is: U = − μ B cos θ = − μ ⋅ B.
U has a minimum value of − μ B for θ = 0 (position of stable equilibrium).
U has a maximum value of μ B for θ = 180° (position of unstable equilibrium).
Note : For both positions the net torque is τ = 0.
                                                                              (28-14)
                    The Hall Effect
        R L         In 1879 Edwin Hall carried out an experiment in which
L                 R
                    he was able to determine that conduction in metals is due
                    to the motion of negative charges (electrons). He was also
                    able to determine the concentration n of the electrons.
                     He used a strip of copper of width d and thickness l. He passed
                     a current i along the length of the strip and applied a magnetic
                     field B perpendicular to the strip as shown in the figure. In the
                             p p                                          g
                                   r                                            r
                     presence of B the electrons experience a magnetic force FB that
                     pushes them to the right (labeled "R") side of the strip. This
    L         R              l t
                     accumulates negative charge on the R-side and l
                                       ti    h         th R id                 th l ft
                                                                      d leaves the left
                     side (labeled "L") of the strip positively charged. As a result
                                                                    r
                     of the accumulated charge, an electric field E is generated as
                     shown in the figure, so that the electric force balances the magnetic
                     force on the moving charges: FE = FB → eE = evd B →
                     E = vd B (eq 1) From Chapter 26 we have: J = nevd →
                               eq. ).
                            J   i    i
                     vd =     =   =            (eq. 2)                    (28-15)
                            ne Ane ldne
L       R L       R        E = vd B (eq. 1)         vd = i / ldne   (eq. 2)
                      Hall measured the potential difference V between the left and
                      the right side of the metal strip: V = Ed (eq. 3).
                      We substitute E from eq. 3 and vd from eq. 2 into eq. 1 and get:
                      V         i                Bi
                         =B         →       n=        (eq. 4)
                      d       ldne              V le
                      Figs. a and b were drawn assuming that the carriers are electrons.
                      In this case if we define V = VL − VR we get a positive value.
                      If we assume that the current is due to the motion
    L         R       of positive charges (see fig. c) then positive charges accumulate on
                      the R-side and negative charges on the L-side, and thus V = VL − VR
                      is now a negative number.
                      By determining the polarity of the voltage th t develops between the
                      B d t      i i th      l it f th        lt   that d l      b t      th
                      left-and right-hand sides of the strip, Hall was able to prove that
                      current was composed of moving electrons. From the value of V
                      using equation 4 he was able to determine the concentration of
    (28-16)           the negative charge carriers.
                             The Cyclotron Particle Accelerator
                             The cyclotron accelerator consists of two hollow
                             conductors in the shape of the letter dee (these are
                             known as the "dees" of the cyclotron). Between the
                                         oscillator
                             two dees an oscillator of frequency f osc creates an
                             oscillating electric field E that exists only in the gap
                               between the two dees. At the same time, a constant
    mv            eB           magnetic field B is applied perpendicular to the
r=           f =
     eB          2π m          plane of the dees.
         g                 y                       p             p
In the figure we show a cyclotron accelerator for protons. The protons follow circular
                     mv                                            eB
orbits of radius r =      and rotate with the same frequency f =        . If the cyclotron
                     eB                                           2π m
                         oscillator
frequency matches the oscillator frequency then the protons during their trip through
the gap between the dees are accelerated by the electric field that exists in the gap.
The faster protons travel on increasingly larger radius orbits. Thus the electric
field changes the speed of the protons while the magnetic field changes only
the direction of their velocity and forces them to move on circular (cyclotron) orbits.
                                                                                  (28-17)