FE Physics

Formulas & Explanation

Formulas & Explanation

Dutch version

**
Electromagnetic Induction
and Alternating Current**

**
**

**
Magnetic
Flux**

**
**

The
magnetic flux is proportional to the number of

field
lines that pass through the surface.

Φ
=
B_{n}
A

Φ
magnetic flux in Wb (Weber). Also
: T.m^{2} of Vs

B_{n}
the component of B perpendicular to the
surface in T(esla)

A
area of the surface in m^{2}

^{
}

*
Example *

See the
drawing above.

B = 1.5
x 10^{-3}
T

A = 500
cm^{2}

α
= 40 ^{
o}

^{
}

Resolve
B in both components and calculate B_{n}

sin
α
= B_{n}/B

B_{n} = B sin 40
^{o
}= (1.5 x 10^{-3 }
)( 0.643) = 9.642
x 10^{-4}
T

Φ
= B_{n}
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:

V_{ind
}
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 cm^{2}

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 90^{o}
. In that situation the magnetic
flux is maximal

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

V_{ind }=-
N ( ΔΦ/Δt)

ΔΦ
= Φ_{max }– 0 = B A = (0.020 )(40 x 10^{-4})^{ }= 8.0 x 10^{-5}
Wb

V_{ind
}= (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

U_{ind }= I_{ind
}
R

1.6 x 10^{-2 }= (I_{ind })(0,20)

I_{ind}
= 8,0.10^{-2 }A

**Sinusoidal
alternating current **

**Peak voltage and
root mean square value of voltage**

V(t) = V_{0} sin
(2πft)
(set calculator to radians)

V voltage in
V

V_{ 0}
_{ }peak
voltage in V

f
frequency in Hz

t
time in s

_{
}

V_{rms }= ½
√2 V
_{0
}

_{
}

V_{rms }
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)

V_{rms }= ½
√2. V_{0
}= 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.

V_{rms }= 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) = V_{0}
sin (2πft)

Correspondingly we can write for
the current:

I(t) = I _{0 }sin (2πft)

I _{rms} =
½ √2.
I_{0}

**
Transformer**

V_{P
} :
V_{S }= N_{P
}: N_{S}

V_{P } voltage
across the primary coil in V

V_{S }voltage
induced in the secondary coil in V

N_{P
}amount of turns in the primary coil

N_{S
}amount of turns in the secondary coil

P_{P }= P_{S
}(Assuming
100 % energy transfer, no energy dissipated in heating))

V_{P }I_{P
}= V_{S }I_{S}

P_{P
}power of the primary coil W (of
VA)

I_{P }
current
in the primary coil in A

P_{S
}power of the secondary coil
in W (of VA)

I_{S
}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

V_{P }
:
V_{S
}= N_{P
}: N_{S}

230 : V_{S} = 400 : 80

230 : V_{S }= 5 : 1

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

V_{S }= 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

(N_{P }: N_{S
} = 5: 1)

Calculate the current in the
primary coil

P_{S }= V_{S
}I_{S}
= (6.0) (0.4) = 2.4 W

V_{P }= 5 V_{S }=
(5) (6.0) = 30
V

P_{P
}= P_{S
}= 2.4 W

P_{P
}= V_{P
}I_{P}

_{
}2.4 = (30)( I_{P}

_{
}I_{P }= 2.4/30 = 0.8 A