Electrical Magnetic Forces
From coulomb's law the magnitude of electrostatic forces between two charges q1 and q2 separated by distance r is given by
F = 1 / 4π?0 q1 q2 / r2 (1)
Consider two parallel conducting wires of length dl1 and dl2 carrying currents I1 and I2 separated by distance.
The magnitude of the magnetic force acting between these two current elements is given by
F = μ0 /4π I1 I2 / r2 dl1 dl2 (2)
let q1 and q2 be the charges flowing for time t in wires for current s I1 and I2. Then
I1 dl1 = q1 / t dl1 = d1 v1 and I2 dl2 = d2 / t dl2 = q2 v2
∴ Fm = μ0 / 4π q1 q2 / r2 v1 v2 (3)
Dividing by we get
Fm / Fe = v1 v2 (μ0 ?0) (4)
Since left hand side is dimensionless therefore right hand side of must also be dimensionless. It means the quantity μo ?o must have th dimensions of (velocity)-2 here v1 and v2 are the drift velocities of electrons incurrent elements As drift velocities are of the order of 10-5ms -1 therefore
v1 v2 = 10-5 X 10-5 = 10 - 10 m2 s-2
It is also found that
C2 = 1 / μ0 ?0 or μ0 ?0 = c-2
Where c is the velocity of light
C = 3 x 108 ms-1
Putting values in (17) we get
Fm / Fe = 10-10 / (3 x 108) 2 = 10-27
It shows that the magnetic forces are very much smaller than the electrostatic forces in current carrying conductors. However, the electric forces never dominate the magnetic forces in current carrying conductors because
The matter is electrically neutral to very high degree of accuracy. The flowing charges do not disturb the neutrality of the matter. Thus the coulomb electric forces are very rarely in evidence.
In an electric current the large numbers of electrons are drifting towards the same direction this adds up weak magnetic fields which becomes evident.
Note if v1 and v2 are comparable to the speed of light the magnetic and electric forces are comparable.
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