# Unit 4 Section 4 Electromagnetic Induction

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## Electromagnetic Induction

An e.m.f. is induced if a conductor is moved through a magnetic field. The conductor can move and the magnetic field can stay still or the other way around - you get an e.m.f. either way.

Flux cutting (i.e. Moving a conductor through a magnetic field) always induces an e.m.f. but will only induce a current if the circuit is complete.

Flux linking is when an e.m.f. is induced by the changing magnitude or direction of the magnetic flux

A change in flux of one weber per second will induce an electromotive force of 1 volt in a loop of wire.

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## Magnetic Flux

Magnetic Flux: The magnetic flux (in Wb) passing through an area is given by the magnetic flux density multiplied by area. It can also be thought of as the number of magnetic field lines passing through an area.

Magnetic flux(OI) = B x A

OI = magnetic flux in Wb (webers)
B = magnetic flux density in T
A = area in m^2

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Flux linkage: The magnetic flux in a coil multiplied by the number of turns on the coil.

Flux linkage (in weber turns) = N x magnetic flux (Wb) = B x A x N

N = number of turns on the coil cutting the flux.

B = magnetic flux density density in Tesla (T)

A = area of the coil in m^2

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E.g.
The flux linkage of a coil with a cross-sectional area of 0.33m^2 normal to a magnetic field of flux density 0.15 T is 4.0 Wb turns. How many turns are in the coil?

Just rearrange the equation for for flux linkage to make N the subject:

1. (N x magnetic flux) = B x A x N

2. N = (N x magnetic flux) / B x A

3. N = 4.0 / (0.15 x 0.33)

4. N = 81 turns

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## Flux Linkage at an Angle

• Magnetic flux, Omega (Wb) = B x A x cos(theta)
• Flux linkage, N Omega in Weber turns  = B x A N x cos(theta)

B = magnetic flux density in Tesla (T)

A = area of the coil in m^2

N = number of turns on the coil cutting the flux

Theta = angle between the normal to the plane of the coil and the magnetic field (degrees)

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## Flux Linkage at an Angle Example

E.g.
A rectangular coil of wire with 200 turns and sides of length 5.00cm and 6.51cm is rotating in a magnetic field with B = 8.56 x 10^-3 T. Find the flux linkage of the coil when it's at 29.5° to the magnetic field.

1. Find the area of the coil:
Area = (5.00 x 10^-2) x (6.51 x 10^-2) = 3.255 x 19^-3 m^2

2. Then just put the numbers into the equation:
N x magnetic flux = BANcos(theta)
= (8.56 x 10^-3) x (3.255 x 10^-3) x 200 x cos(29.5°)
= 4.85 x 10^-3 Wb turns

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