Physics Laboratory 2
last update: 2009. 8. 31
Exp. 4. What’s Happening Between Currents?
- Lorenz Force -
Purpose of Experiment
A current-carrying electric wire produces a magnetic field. When a closed currentcarrying wire is placed in a magnetic field, it feels a torque. If the magnetic field is not
uniform (that is, the magnitude and direction depend on the position) such as near a
straight wire, two wires exert forces on each other. In this experiment, we measure the
torque acting on an electric wire which is placed in a magnetic field produced by a
solenoid with dc current. We determine the direction of the torque and see the
dependence of the magnitude of the torque on the magnetic field and the current
flowing in the wire.
- From E. Hecht, Physics
An electric motor would be a typical example of practical use of the torque that a current (or
current-carrying wire) experiences in a magnetic field. But an amperemeter (or voltmeter) we
frequently use in the laboratory also employs the torque in uniform magnetic field as left picture.
And a speaker also employs the torque acting on the coil in nonuniform magnetic field as right
picture. What would be the direction of the magnetic field? Pay attention to the position of
speaker's coil.
Outline of Experiment

Confirm a closed current-carrying electric wire experiences a torque in a uniform
magnetic field.

Examine the characteristics of a magnetic torque exerting on a wire
- How does it depend on the magnitude and the direction of a magnetic field?
- How does it depend on the current in wire?

Understand the principle of a current balance.
Experimental Method
These equipments are prepared in the laboratory. (Parentheses mean the number of
them.)
solenoid (500 turns, 1)
dc dual power supply (1)
electric wire with crocodile clip (6)
variable resistor (rheostat) (1)
current balance (1)
30 cm ruler (1)
digital scale (1, common use)
micrometer (2, common use)
Scotch Tape (2, common use)
cutter knife (2, common use)
If you need more stuff, inquire to your teaching assistant or experiment preparation
room (19-114), or prepare yourself. Learn in advance about the current balance.
The following is a recommended experiment method.
1) Connect the solenoid to a port of the dual power supply. [movie : experiment 2]
2) Confirm the power supply's current handle is turned counterclockwise. Then turn on
the power supply. Rotate the handle until the amperemeter displays 1A.
3) Put the current balance into the solenoid. The current-flowing portion should be in.
Don't let the inner part touch the inner wall.
4) Place the indicator on the outside of the balance. Modulate the weight control screw
so that the balance tilts up or down slightly but not touch the inner wall and be in
equilibrium.
5) Connect the support screw of the balance with another port of dual power supply and
the variable resistor (rheostat) in series to flow current.
6) Rotate the current control handle of a port connected with balance until the
amperemeter displays about 1A.
7) While turning off and on, see the balance tilt toward the opposite direction. If not,
change the polarity of the current in the solenoid or in the balance.
8) For 9 values of the current in balance in 0 ~ 2 A, measure the position of indicator
using 30cm ruler. [video : experiment 3]
9) Repeat 8) for 5 values for the current in solenoid in 0 ~ 2.5 A.
[caution 1 : The currents in balance and in solenoid should not over 2 A and 2.5 A
respectively. Because a current above 2 A could heat the wire, don't flow it overlong.]
[caution 2 : Don't detach the balance from the support screw while a current flows. (The
spark could cause some damage.)]
[reference : variable resistor (rheostat) - a resistor which can vary the resistance using
control handle. It has a control connector and operating handle. The operating handle
moves the position of resistance which changes the resistance between connectors. In
this experiment, a variable resistor is used to reduce the current in balance to 0. It can't
be reduced to 0 without it.]
10) How does the force acting on a current-carrying wire in a magnetic field depend on
the current in wire? How about on the magnetic field? Is it possible to compare the
experimental result with the theoretical calculation?
Below is an example form of experiment note.
(Measurement of displacements of current balance)
Turn density of solenoid m =
/m
Width of current balance d =
m
Length from center line of balance to copper(wire) l =
m
Length from center line of balance to the tip of indicator s =
Mass of balance m =
kg
Radius of support axis of balance r =
Current in solenoid I =
Current in balance I'(A)
0
0.3
0.5
0.75
1.0
1.25
m
m
A
Position of ruler y(m)
Displacement of indicator △y(m)
1.5
1.75
2.0
Slope △y/I' =
m/A
Slope △y/I'(m/A)
Current in solenoid
0.5
1.0
1.5
2.0
2.5
m/A2
Slope △y/(II') =
Background Theory
A force between currents in separated wires can be understood as one current feels a magnetic
field induced by another current. Let the currents in parallel wires be ia and ib respectively and
the distance between them d. The magnitude of magnetic field induced by ia is
Ba = μ0ia/(2πd) (1)
at the position current ib flows and the direction is downward by the right hand rule. μ0 is the
permeability which is defined as μ0 = 4π×10-7T·m/A in vacuum or matter with no magnetism. It
will be explained later again.
The force exerted on length L of a wire b with current ib is
→
→ →
Fba = ibL × Ba
(2)
According to the picture, the magnetic field and the current ib is mutually perpendicular, so that
the magnitude of the magnetic field is
Fba = ibLB = μ0 iaibL/(2πd) (3)
and the force per unit length is
Fba/L = μ0 iaib/(2πd) (4)
Since the force is measurable, for two wires carrying the same current i separated by 1m, we
define 1 A (ampere) as the magnitude of the current i which will produce the force of 2×10-7 N.
Then the proportionality constant in equation (4) (permeability) is
μ0 = 4π×10-7N/A2
(5)
or using
1 N = 1 A·m·T
(6)
, it becomes
μ0 = 4π×10-7T·m/A (7)
When a loop carrying current I is inclined at φ in a uniform magnetic field such as in an ideal
solenoid, it experiences a torque by the magnetic field.
In the picture, for horizontal length a, vertical length b and turn number N=1, the magnitude of
the torque is given by
τm = IabBsinφ
(8)
In an ideal solenoid, the magnetic field is parallel to the axis of solenoid and the magnitude is
given by
Bsolenoid = μ0isolenoidn
(9)
where n is the turn per unit length, namely turn density of solenoid. In reality, since the length of
a solenoid is finite, the magnetic field exists outside the solenoid and the inside field is also
slightly different from ideal case. For a solenoid of length l and radius r, the magnetic field at the
center is
Bsolenoid = μ0isolenoid n cosθc
(10)
where θc is the angle between the axis and the straight line from the center of solenoid to the end.
In this case,
cosθc = (l/2){r2+(l/2)2}-1/2 (11)
When a loop is partially included in the magnetic field like this experiment, the torque acting on
the loop is
τm = (1/4)μ0Iabisolenoidnl{r2+(l/2)2}-1/2sinφ
(12)
The current balance uses the equilibrium between the magnetic torque above and the restoring
torque exerted by the balance.
The force exerted by the magnetic field is not only the case of the torque acting on a currentcarrying wire. In fact, there were discussions about whether a magnetic force acts on a wire or a
current itself. For now, we understand it as charges which carry the current feel the force first
and then almost simultaneously the wire feels it since they can't get out of it. When a particle
with the charge q moves at the velocity v in a magnetic field B, this particle experiences the
magnetic force of
→
→ →
F = q v × B (13)
This is called Lorentz force. With an electric field E, Lorentz force denotes, including the electric
force,
→
→
→ →
F = q E + q v × B (14)
The first example of Lorentz force is the discovery of electron by J. J. Thomson, which is still
used to control the direction of electron beams in Braun tube in computer monitor (CRT,
Cathode Ray Tube)
From E. Hecht, Physics
In cathode ray tube (CRT), to deflect the electron beams one can use the electric field (left
picture) or the magnetic field. The method using the magnetic field which is employed in home
televisions and computer monitors uses the yoke coil (right picture) instead of the deflection
electrode. If it has tiny pixel size like a natural color television, very tiny magnetic field from
Earth or nearby magnetic substances could affect on its color system. Indeed, the exported
televisions, depending on Earth's magnetic field at their destination, are corrected in advance.
Things to Figure Out

Current Balance
A current balance uses the mechanical equilibrium condition between the magnetic torque a wire
feels and the torque by gravitational force. The current balance used in this experiment is an
electric circuit board coated by copper, which can rotate around its center axis. Note that the
support axis which also does a role of the electrode wire is attached on the upper side of the
circuit. It gives the stability so that the balance can do its roles.
Now, we will see how a current balance keeps up the stability. Assume the balance is in
equilibrium when no current flows. When the solenoid applies a magnetic field B, the balance
feels a torque τm, so it starts to rotate. But for the balance where the support is attached on the
upper side, when the support is rotated, the perpendicular line passing circuit's center of mass
doesn't pass the contact point of the support. Thus the restoring torque by gravity acts on the
circuit. Let the mass of the balance be M, the radius of support be r and the gravitational
acceleration be g. Then the restoring torque is
τ = Mgrsinθ
(A1)
, so that the torque increases as the angle θ increases. When it compensates for the magnetic
torque, the circuit becomes balanced. In the equilibrium, the relation between the torque and the
angle is given by
τm = Mgrsinθ ≒ Mgrθ (A2)
If the angle is small, they are mutually proportional.
In this experiment, we calculate the angle from the displacement of indicator attached on the
balance. Let the distance from the support to the indicator be L and the vertical displacement of
the indicator be y. Then the angle of circuit is
θ = tan-1(y/L) ≒ y/L
(A3)
and the torque acting on the circuit is
τm ≒ Mgry/L (A4)
Note that we have assumed the circuit is negligibly thin and uniform and it becomes horizontal in
the equilibrium state with no magnetic field.
References

Treatment of measurement data

Analysis method based on the graph

Sir Joseph John Thomson - An excellent trainer who found the electron.

Robert Millikan - Pride of modern American physics
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