Download A homopolar motor

A homopolar motor
Figures 1 a) and b), taken from Wikipedia,1 show
the design of an “homopolar motor”. The idea is
the following: the battery drives a DC current into
the circuit and the magnetic field generated by the
magnet at the bottom thus generates a torque on
the circuit, which starts to rotate.2
Describe how the motor works. In particular,
assuming the dimensions, mass and resistance of
the circuit, the voltage of the battery and the magnetic field strengh of the magnet near its surface to
be known, find the torque on the circuit and the
angular velocity as a function of time.
a)
b)
Figure 1
1
The figures are in the public domain. See http://en.wikipedia.org/wiki/File:Motor_homopolar.png and
http://en.wikipedia.org/wiki/File:Motor_homopolar_flux_force.png.
2
A video showing how a similar motor works can be found at http://www.youtube.com/watch?v=kH52QbLxhGs.
1
Solution
The motor can be represented by the schematic in Figure 2,
G
F
where thanks to the symmetry we may keep only “half” of
+
the circuit. The generator V represents the battery. The
V
rectangular circuit CGF ED is connected to a thin conduct−
B
ing disk by a contact at C (at the center of the disk), and
h
a sliding contact at D (at the periphery of the disk): the
electrical circuit is closed by the current flowing in the disk.
The latter is placed over the top base of the cylindrical magnet. Let a and h be the horizontal and vertical lengths of the
C
E
D
circuit, respectively and b < a the radius of the disk and the
magnet. Let B = B(r, z) be the magnetic field with cylinb
drical symmetry generated by the magnet (some field lines
a
of B are sketched in Fig.1 b), and B0 be the magnetic field
strength at the surface of the magnet (i.e. B0 = Bz (r, 0) for
0 < r < b). The circuit can rotate freely around the z = 0
Figure 2
axis that passes through C and G.
The voltage generator drives a constant current I in the circuit (we may
dl θ
argue at this point that I = V /R but this will turn out not to be correct
B
because
of the induced electromotive force; this point will be addressed later).
n̂
r
Since the magnetic field B is always parallel to the circuit, the force df =
I
−IB × dl on an infinitesimal segment dl is perpendicular to the plane of
the circuit. This gives rise to an infinitesimal torque dM = −Ir × (B × dl)
parallel to the axis, where r is the distance of dl from the axis itself. Notice that we can write
B × dl = (|B||dl| sin θ)φ̂ = (B · ndℓ)φ̂ where n is the unit vector perpendicular to dl and dℓ = |dl|.
Since r is perpendicular to φ̂, the total modulus of the torque is given by the integral over the circuit
(notice: we do not keep track of the signs!)
Z
Z
Z
Z
Z |M| =
IB · nrdℓ = I
+
+
+
B · nrdℓ
CGF ED
CG
GF
FE
ED
Z a
Z a
Z h
Bz (r, 0)rdr ,
(1)
= I
Br (a, z)adz +
Bz (r, h)rdr +
b
0
0
since the integral vanishes over CG, where Br = 0.
To evaluate |M| let us first demonstrate that extending the integration over the closed path the
integral vanishes:
I
Z
Z b
Bz (r, 0)dr = 0 .
(2)
B · nrdℓ =
B · nrdℓ +
CGF EDC
CGF ED
0
In fact, this expression is proportional to the flux of B through the closed cylindrical surface spanned
by the rotation of the circuit:
Z a
Z
Z a
Z h
0 = Φ(B) = B · ndS = 2π
Bz (r, 0)rdr .
(3)
Br (a, z)adz +
Bz (r, h)rdr +
0
0
2
0
We thus obtain
Z
B · nrdℓ = −
CGF ED
Z
b
Bz (r, 0)rdr ≃ −B0 b2 /2 ,
(4)
0
where we assumed the magnetic field at the surface of the magnet to be uniform.
This description should clarify the essentials of the motor, in particular why we need sliding
contacts and a conducting “cap” of the magnet instead of simply taking a closed rectangular circuit
free to rotate along the axis. In the latter case, the torque produced by the forces drawn in Fig.1 b)
would be canceled out exactly by the torque due to the forces near the surface of the magnet, i.e. at
z = 0 and for b < r < a. Thus, putting a sliding contact at the edge of the magnet allows to achieve
the maximum torque since we keep only “positive” contributions of the magnetic force all over the
circuit.
The torque M = IB0 b2 /2 makes the circuit rotate with angular velocity ω such that Idω/dt = M ,
where I is the momentum of inertia. Thus, keeping the circuit size, geometry and resistance as fixed,
we may increase ω by choosing a stronger magnet. However, increasing ω should lead to a higher
dissipated power P = RI 2 provided by the battery. Thus, there must be an electromotive force
contribution E such that the current I = (V − E)/R depends on ω. Such electromotive force may be
calculated as the integral of the v × B force along the rotating circuit, where v is the local velocity
(this is one of the cases in which it the electromotive force is not simply given by the variation of the
magnetic flux across the circuit, which is always zero since B is parallel to the plane of the circuit
itself). We may also state that the electromotive force originates from the flowing of the current
along the conducting disk against the magnetic field. This suggests that the value of E is more easily
estimated by considering a rotating frame where the circuit is at rest, hence the disk rotates in the
magnetic field. Thus this is a Faraday disk generator, which generates a potential drop along its
radius
Z
Z b
rdr = B0 ωb2 /2 .
(5)
E = (r × ω) × B · dr = B0 ω
0
We thus obtain
I=
V − B0 ωb2 /2
,
R
(6)
so that
dω
=
I
dt
V − B0 b2 ω/2
R
B 0 b2
,
2
(7)
with the solution
ω=
2V
1 − e−t/τ ,
2
B0 b
τ=
4RI
.
(B0 b2 )2
(8)
Putting in some “reasonably feasible” values such as V = 1.5 Volts, B0 = 100 Gauss, and b = 0.5 cm
yields impressively high values of ω! It is much more likely that friction effects determine the
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asymptotic angular velocity keeping it at a much lower value. Nevertheless the inclusion of the
electromotive term is necessary to account for higher dissipation in the case of a stronger magnetic
field.
Thanks to Alessandro Porcelli3 for having stimulated this discussion, which originated from the
attempt to build his own homopolar motor.
3
[email protected]
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