MadSci Network: Physics Query:

### Re: Is there any change to a Magnetic field as it gInduces a current flow ?

Date: Fri Jun 8 17:34:06 2001
Posted By: Yaxun Liu, Grad student, Electrical and Computer Engineering, University of Waterloo
Area of science: Physics
ID: 991394550.Ph
Message:
```
This is a good question. The answer depends on what you
connect to the two leads of the coil. If you let it open,
there will be no current induced in the coil, therefore
no effect on the magnetic field. If you short-circuit
the two leads, and the resistance of the coil is very
small, there will be a very strong induced magnetic field
which will almost cancel out the imposed magnetic field.

Consider the magnetic field penetrating the coil. It is
composed of two parts: one is the external magnetic field B_e
which does not depend on the current in the coil, another
is the internal magnetic field B_i caused by the current I
in the coil.

A useful quantity related to this problem is the magnetic
flux. If you view magnetic field as a flowing stream, the
magnetic flux is the quantity of magnetic field flowing
through the coil.

The magnetic flux F_i due to B_i is proportional to the
current I in the coil,

F_i = I * L

Actually L is just the inductance of the coil.

Denote the magnetic flux due to B_e as F_e, there will
be a voltage induced at the two leads of the coil, which is

d                    dI   dF_e
V =  -- (F_i + F_e) = L * -- + -----
dt                   dt   dt

Therefore, if you draw a equivalent circuit for the coil, it
is a inductance L in series with a voltage source Vs = dF_e/dt.

For a current to appear in the coil, you need to complete the
circuit by connecting the two leads of the coil with a wire or
a resistor or a capacitor or anything which can let a current pass
through it. The the current will be determined by the whole circuit.

For example, if the external magnetic field is a sinusoidal
function of time with angular frequency as w, then you will
get a sinusoidal voltage source Vs = jw*F_e (Here for simplicity
we use phasor to represent the quantities and assume the time
dependence as exp(j*w*t), j is the square root of -1).

Assume you connect a resistor R at the two leads of your coil,
then the current in the coil will be

-Vs    -jw*F_e
I = ----- = --------
jwL+R    jwL+R

The total magnetic flux through the coil is

F_i + F_e = L * I + F_e = R * F_e / (jwL + R)

Therefore, if you let the coil open-circuited, that is, you
connect nothing to its two leads, the magnetic field through
the coil will not be affected by the coil, since there is no
current induced in it. However, If you connect the two leads
of the coil through a wire, and the coil and the wire are
very good conductance, the magnetic field caused by the
induced current will be so strong that it will almost totally
cancel out the imposed external magnetic field.

Some tutorials on phasors: http://www.clarkson.edu/~svoboda/eta/phasors/Phasor10.html http://www.clarkson.edu/~svoboda/eta/phasors/MatchPhasors10.html http://www.clarkson.edu/~svoboda/eta/phasors/AddPhasors10.html

```

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