FE Physics
Formulas & Explanation     Electromagnetic Induction and Alternating Current

Magnetic  Flux The magnetic flux is proportional to the number of

field lines that pass through the surface.

Φ = Bn A

Φ   magnetic flux in Wb (Weber).  Also :  T.m2 of Vs

Bn   the component of B perpendicular to the  surface in T(esla)

A     area of the surface  in m2

Example

See the drawing above.

B = 1.5 x 10-3 T

A = 500 cm2

α = 40 o

Resolve B in both components  and calculate Bn

sin α = Bn/B

Bn = B sin 40 o = (1.5 x 10-3 )( 0.643) = 9.642 x 10-4  T

Φ = Bn A

Φ = (9.642 x 10-4 )( 500 x 10-4 )= 4.82 x 10-5 Wb

Change of magnetic flux and electromagnetic induction

A change in the magnetic flux in a closed circuit (coil)

induces  a potential difference .

Law of Faraday: Vind  average induced potential difference in V

N      amount of loops of the coil

ΔΦ   change of magnetic flux in Wb

Δ t    time interval in which change occurs in s

Example See the figure alongside.

The area of the turn is  40 cm2

The magnitude of the magnetic induction is

0.020 T

The turn is rotated.

At the moment of the figure the field lines are parallel to  the turn.

So the magnetic flux is 0 Wb

After  0.0050 s the turn has rotated  90o . In that  situation the magnetic flux is maximal

Determine the average induced potential difference (emf) in this time interval.

Vind =- N ( ΔΦ/Δt) ΔΦ = Φmax – 0 = B A = (0.020 )(40 x 10-4) = 8.0 x 10-5 Wb

Vind = (1) (8.0 x 10-5)/0.0050 = 1.6 x 10-2 V

An induction current is produced when A is connected to B

The resistance of the circuit is 0.20 Ω

Uind = Iind R

1.6 x 10-2 = (Iind )(0,20)

Iind = 8,0.10-2 A

Sinusoidal alternating current

Peak voltage and root mean square value of voltage

V(t) = V0  sin (2πft)        (set calculator to radians)

V         voltage in V

V 0       peak voltage in V

f          frequency in Hz

t          time in s

Vrms = ½ 2 V 0

Vrms     root-mean-square value of voltage

Example 1

In the drawing aside:

V 0 =10 V

T = 50 ms

f = 1/T=1/(50 x 10-3) = 20 Hz

So :

V(t) = 10 sin (2π 20 t)

Vrms = ½ 2. V0 = ½ 2. 10 = 7.0 V

This means :

A  DC voltage of  7.0 V produced the same power as an AC voltage  with a peak voltage of

10.0 V

Example 2

The mains voltage is :  V =230 V  (This means :  V rms = 230 V)

Find the peak voltage.

Vrms = ½ 2  V 0

230 = ½ 2  V 0

V 0 = 325 V

Peak current and root-mean-square value of the current

In the previous paragraph we see : V(t) = V0  sin (2πft)

Correspondingly  we can write for the current:

I(t) = I 0 sin (2πft)

I rms = ½ 2. I0

Transformer VP  :  VS  = NP :  NS

VP   voltage across the primary coil in V

VS    voltage induced in the secondary coil in V

NP    amount of turns in the primary coil

NS    amount of turns in the secondary coil

PP = PS                                   (Assuming 100 % energy transfer, no energy dissipated in heating))

VP IP = VS IS

PP    power of the primary coil W (of VA)

IP     current in the primary coil in A

PS      power of the secondary coil  in W (of VA)

IS       current  in the secondary coil  in A

Examples

1.         The primary coil of a transformer consists of 400 turns. The secondary coil

80 turns.

The voltage across the primary coil is 230 V

Find the voltage across the secondary coil

VP  :  VS  = NP :  NS

230 : VS = 400 : 80

230 : VS = 5 : 1

(5)(Vs) = (230) (1)

VS = 230/5 = 46 V

2.           The primary coil is connected with an AC source

A bulb is connected with the secondary coil.

The voltage across the bulb is 6.0 V and

the current through the bulb is  0.4 A

The turns ratio is 5:1

(NP : NS  = 5: 1)

Calculate the current in the primary coil PS = VS IS = (6.0) (0.4) = 2.4 W

VP = 5 VS = (5) (6.0) = 30 V

PP = PS = 2.4 W

PP = VP IP

2.4 = (30)( IP)

IP = 2.4/30 = 0.8 A