Experiment 4f
Class:
Name:
(
) Date:
4f Effect of magnetic field strength on
magnetic force
Objective
To find out how the magnetic force on a current-carrying conductor
is affected by the magnetic field strength.
Background information
1
The magnetic field strength at the centre of an infinitely long currentcarrying solenoid is directly proportional to the current through it. It
is given by the equation: B = µ 0nI.
2
When a current flows through a conductor in a magnetic field across
it, a magnetic force acting on it is produced.
3
The graph in Figure 4f-1 is a straight line
passing through the origin, which means x
and y are directly proportional, i.e.
y∝x
or y = kx
(k is a constant)
y
y0
slope = x = k
0
y0
origin
x0
x
Fig 4f-1
Apparatus
❏ 1 current balance
❏ 2 power packs (0–12 V a.c./d.c.)
❏ 1 flat solenoid
❏ 1 electronic balance
❏ 2 ammeters
❏ 2 rheostats
❏ several connecting leads
70
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© Oxford University Press 2007
Class:
Name:
(
Experiment 4f
) Date:
Procedure
Note
1
(a) Set up a current balance (Fig 4f-2).
(b) Adjust the position of the wire until it is balanced.
(c) Set the reading of the electronic balance to zero.
The reading is in gram
for some electronic
balances. Since an object
of 1 gram has a weight
of 0.01 N on earth, the
reading can be changed
to values of force easily
by multiplying it by 0.01.
power pack
rheostat
ammeter
✐ In order to obtain directly
proportional relationship
of F and B, the arm of the
current balance must be
horizontal and sit lightly on
the plate of the electronic
balance. Some books of
suitable thickness may be
necessary to achieve this.
flat soleniod
insulator
electronic
balance
power pack
ammeter
rheostat
Fig 4f-2
2
(a) Connect a flat solenoid in series with a rheostat, an ammeter
and a d.c. power supply.
(b) Place the current-carrying arm of the current balance inside the
flat solenoid.
3
(a) Pass a fixed current of 3 A through the wire of the current
balance.
(b) Switch on the power supply connecting to the flat solenoid.
Vary the current flowing through the solenoid by changing the
resistance of the rheostat.
(c) Take the reading of the ammeter connecting to the flat solenoid
and the reading of the electronic balance. Record the results in
Table 4f-1 on p.72.
✐ The magnetic field
produced by the solenoid
is quite small. Therefore,
the current through the
wire should not be too
small, say 1 A, otherwise
the magnetic force
produced will be too small.
Note
The reading of the
electronic balance gives
the magnetic force acting
on the current-carrying
arm.
New Physics at Work (Second Edition)
© Oxford University Press 2007
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Experiment 4f
Class:
Name:
(
) Date:
4
Repeat several times with other current values through the flat
solenoid. Record the results in Table 4f-1.
✎
Results:
Current through flat solenoid / A
(∝ magnetic field strength B)
0.5
1.0
1.5
Magnetic force on wire F / × 10–3 N
0.6
1.3
1.8
Table 4f-1
5
Plot a graph of the magnetic force F on wire against the current
through flat solenoid (directly proportional to the magnetic field
strength B) in Figure 4f-3.
magnetic force
on wire F / t 10–3 N
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0
0.5
1.0
1.5
2.0
2.5 current through flat solenoid / A
(u magnetic field strength B)
Fig 4f-3
72
New Physics at Work (Second Edition)
© Oxford University Press 2007
Class:
Name:
(
Experiment 4f
) Date:
Discussion
✎
How is the magnetic force F on the wire related to the magnetic field
strength B?
The magnetic force on the wire is directly proportional to the magnetic field strength,
i.e. F ∝ B.
The magnetic force on a current-carrying conductor is
directly proportional
_______________________________ to the magnetic field strength.
New Physics at Work (Second Edition)
© Oxford University Press 2007
73