## AQA A Level Physics

### Revision Notes

• More coils in a wire mean a larger e.m.f is induced
• The magnetic flux linkage is a quantity commonly used for solenoids which are made of N turns of wire
• The flux linkage is defined as:

The product of the magnetic flux and the number of turns of the coil

• It is calculated using the equation:

Magnetic flux linkage = ΦN = BAN

• Where:
• Φ = magnetic flux (Wb)
• N = number of turns of the coil
• B = magnetic flux density (T)
• A = cross-sectional area (m2)

• The flux linkage ΦN has the units of Weber turns (Wb turns)

### Flux Linkage in a Rotating Rectangular Coil

• When the magnet field lines are not completely perpendicular to the area A, then the component of magnetic flux density B is perpendicular to the area is taken
• The equation then becomes:

Φ = BA cos(θ)

• Where:
• Φ = magnetic flux (Wb)
• B = magnetic flux density (T)
• A = cross-sectional area (m2)
• θ = angle between magnetic field lines and the line perpendicular to the plane of the area (often called the normal line) (degrees)

The magnetic flux through a rectangular coil decreases as the angle between the field lines and plane decrease

• This means the magnetic flux is:
• Maximum = BA when cos(θ) =1 therefore θ = 0o. The magnetic field lines are perpendicular to the plane of the area
• Minimum = 0 when cos(θ) = 0 therefore θ = 90o. The magnetic fields lines are parallel to the plane of the area
• An e.m.f is induced in a circuit when the magnetic flux linkage changes with respect to time
• This means an e.m.f is induced when there is:
• A changing magnetic flux density B
• A changing cross-sectional area A
• A change in angle θ
• As with magnetic flux, if the field lines are not completely perpendicular to the plane of the area they are passing through
• Therefore, the component of the flux density which is perpendicular is equal to:

ΦN = BAN cos(θ)

• Where:
• N = number of turns of the coil

#### Worked Example

A solenoid of circular cross-sectional radius 0.40 m2 and 300 turns is placed perpendicular to a magnetic field with a magnetic flux density of 5.1 mT.

Determine the magnetic flux linkage for this solenoid.

Step 1: Write out the known quantities

• Cross-sectional area, A = πr2 = π(0.4)2 = 0.503 m2
• Magnetic flux density, B = 5.1 mT
• Number of turns of the coil, N = 300 turns

Step 2: Write down the equation for the magnetic flux linkage

ΦN = BAN

Step 3: Substitute in values and calculate

ΦN = (5.1 × 10-3) × 0.503 × 300 = 0.7691 = 0.8 Wb turns (2 s.f)

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