Experiment 1
Connect a small electric motor to a moving-coil voltmeter (Figure). Spin the shaft of the motor and observe the deflection of the voltmeter.
What happens when you spin the motor more slowly?
What happens when you stop? Usually, we connect a motor
to a power supply and it turns. In this experiment, you have turned the motor and it generates a voltage across its terminals. A generator is like a motor working in reverse.
Experiment 2
Connect a coil to a sensitive microammeter (Figure).
Move a bar magnet in towards the coil. Hold it still, and then remove it. How does the deflection on the meter change? Try different speeds, and the opposite pole of the magnet. Try weak and strong magnets.
With the same equipment, move the coil towards the magnet and observe the deflection of the meter.
Experiment 3
Connect a long wire to a sensitive microammeter. Move the middle section of the wire up and down through the magnetic field between the magnets (Figure).
Double up the wire so that twice as much of it passes through the magnetic field. What happens to the meter reading? How can you form the wire into a loop to give
twice the deflection on the meter?
Discussion and demonstration: Induction effects
The first two demonstrations involve moving a wire in a magnetic field and then a permanent magnet into and out of a small coil. In both it is important to emphasise that:
‘electricity’ is only produced while something is moving
the faster the movement, the more ‘electricity’ we get
Cutting magnetic field lines
What happens when a wire is moved into the magnetic field?
As it moves, it cuts across the magnetic field. Remove the wire from the field, and again it must cut across the field lines, but in the opposite direction.
For a coil of N turns, the effect is N times greater than for a single turn of wire.
When the coil is outside the field, there are no magnetic field lines linking the coil.
When it is inside the field, field lines link the coil. Moving the coil into or out of the field changes this linkage, and this induces an e.m.f. across the ends of the coil.
Magnetic flux and magnetic flux linkage
Magnetic flux density B is defined by the equation
B = F/IL
Now we can go on to define magnetic flux as a quantity.
We picture magnetic flux density B as the number of
magnetic field lines passing through a region per unit area.
Similarly, we can picture magnetic flux as the total
number of magnetic field lines passing through an area A. For a magnetic field normal to A, the magnetic flux Φ
must therefore be equal to the product of magnetic flux density and the area A.
The magnetic flux Φ through area A is defined as:
Φ = BA
where B is the component of the magnetic flux density
perpendicular to the area.
How can we calculate the magnetic fl ux when B is not
perpendicular to A?
When the field is parallel to the plane of the area, the magnetic flux through A is zero. To find the magnetic flux in general, we need to find the component of the magnetic flux density perpendicular to the area.
Magnetic flux = (B cos θ) × A
or simply:
Magnetic flux = BA cos θ
(Note that, when θ = 90°, flux = 0 and when θ = 0°,
flux = BA.)
For a coil with N turns, the magnetic flux linkage is defined as the product of the magnetic flux and the
number of turns; that is:
Magnetic flux linkage = NΦ
or
Magnetic flux linkage = BAN cos θ
One weber (1 Wb) is the flux that passes through an area of 1 m2 when the magnetic flux density is 1 T.
1 Wb = 1 T m2.
The unit for magnetic flux or flux linkage is the Weber (Wb).
An e.m.f. is induced in a circuit whenever there is a change in the magnetic flux linking the circuit. Since magnetic flux is equal to BA cos θ, there are three ways an e.m.f. can be induced:
■■ changing the magnetic flux density B
■■ changing the area A of the circuit
■■ changing the angle θ.
(W) Whole class work
Figure shows a solenoid with a cross-sectional area 0.10 m2. It is linked by a magnetic field of flux density 2.0 ×10−3 T and has 250 turns. Calculate the magnetic flux and flux linkage for this solenoid.
Step 1 We have B = 2.0 ×10−3 T, A = 0.10 m2, θ = 0°
and N = 250 turns. Hence we can calculate the flux Φ.
Φ = BA
Φ = 2 .0 × 10−3 × 0.10 = 2.0 × 10−4 Wb
Step 2 Now calculate the flux linkage.
magnetic flux linkage = NΦ
magnetic flux linkage = 2.0 ×10−4 × 250
= 5.0 × 10−2 Wb
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