MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2003
VIII. Magnetic Fields - Worked Examples
Example 1: Rolling rod
A rod with a mass m and a radius R is mounted on two parallel rails of length L separated
by a distance d, as shown in the figure below.
The rod carries a current I and rolls without slipping along the rails which are placed in a
G
uniform magnetic field B (directions shown in the figure). If the rod is initially at rest,
what is its speed as it leaves the rails?
Solution:
Using the coordinate system shown on the right, the
magnetic force acting on the rod is given by
G
G G
FB = Id × B = I (d ˆi ) × (− Bkˆ ) = IdBˆj
(1.1)
The total work done by the magnetic force on the rod
as it moves through the region is
G
G
W = ∫ FB ⋅ d s = FB L = IdBL
(1.2)
By the work-energy theorem, W must be equal to the change in kinetic energy:
0
∆K =
1 2 1 2
mv + I ω
2
2
(1.3)
where both translation and rolling are involved. Since the moment of inertia of the rod is
given by I = mR 2 / 2 , and the condition of rolling with slipping implies ω = v / R , we
have
2
IdBL =
1 2 1  mR 2   v  1 2 1 2 3 2
mv + 
   = mv + mv = mv
2
2  2  R 
2
4
4
(1.4)
Thus, the speed of the rod as it leaves the rails is
v=
4 IdBL
3m
(1.5)
Example 2: Magnetic dipole moment in B field
G
G
A current loop with magnetic dipole moment µ is placed in a uniform magnetic field B .
Show that its potential energy is given by
G G
U = −µ ⋅ B
(2.1)
Solution:
G G G
The magnetic field exerts a torque τ = µ × B of magnitude τ = µ B sin θ on the dipole,
tending to turn the dipole moment in the direction of decreasing θ .
We can choose the potential energy U to be zero when the dipole moment is at an angle
θ = π / 2 to the magnetic field. Thus, the energy is given by
U −0 = ∫
θ
π /2
or we can write
µ B sin θ dθ = µ B [ − cos θ ]π / 2 = − µ B cos θ
θ
G G
U = −µ ⋅ B
(2.2)
(2.3)
The potential energy U represents the amount of work which needs to be done by an
G
G
external agent to orient the dipole µ at an angle θ with the magnetic field B .
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Example 3: Suspended conductor
Suppose a conductor having a mass density λ kg/m is suspended by two flexible wires in
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a uniform magnetic field Bin which points into the page (see figure below).
If the tension on the wires is zero, what are the magnitude and the direction of the current
in the conductor?
Solution:
In order that the tension in the wires be zero, the magnetic
G
G G
force FB = IL × B acting on the conductor must exactly
G
cancel the downward gravitational force Fg = −mgkˆ .
G
G
For FB to point in the +z direction, we must have L = Lˆj , i.e., the current flows to the
right, so that
G
G G
FB = IL × B = I ( Lˆj) × (− Bˆi ) = − ILB(ˆj × ˆi ) = + ILBkˆ
(3.1)
The magnitude of the current can be obtain from
ILB = mg
(3.2)
or
I=
mg λ g
=
BL
B
(3.3)
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Example 4: Charged particles in B field
Particle A with charge q and mass mA and particle B with charge 2q and mass mB, are
accelerated from rest by a potential difference V , and subsequently deflected by a
uniform magnetic field into semicircular paths. The radii of the particle A and B are R and
2R, respectively. The direction of the magnetic field is perpendicular to the velocity of
the particle. What is their mass ratio?
Solution:
The kinetic energy gained by the charges is equal to
1 2
mv = qV
2
(4.1)
which yields
v=
2qV
m
(4.2)
The charges move in semicircle, since the magnetic force points radially inward and
provides the source of the centripetal force:
mv 2
= qvB
r
(4.3)
The radius of the circle can be readily obtained as:
r=
mv m 2qV 1 2mV
=
=
qB qB
m
B
q
(4.4)
which shows that r is proportional to (m / q )1/ 2 . The mass ratio can then be obtained from
rA (mA / q A )1/ 2
=
rB (mB / qB )1/ 2
⇒
(mA / q )1/ 2
R
=
2 R (mB / 2q )1/ 2
(4.5)
which gives
mA 1
=
mB 8
(4.6)
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Example 5: Bar magnet in non-uniform magnetic field
A bar magnet with its north pole up is placed along the symmetric axis below a horizontal
conducting ring carrying current I, as shown in the figure below.
What is the force on the ring?
Solution:
G
The magnetic force acting on a small differential current-carrying element Id s on the
G
G
G G
ring is given by dF = Id s × B , where B is the magnetic field due to the bar magnet. Due
to the axial symmetry, the magnetic field lines will intersect the loop at right angles
(figure below), so that the x component of the force will exactly cancel.
G
Thus, net force experienced by Id s is
dFy = dF sin θ = ( IdsB) sin θ
(5.1)
The total force acting on the ring then becomes
Fy = v∫ dFy = IB sin θ v∫ ds = 2π rIB sin θ
(5.2)
The force points in the +y direction and therefore is repulsive.
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