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© The scientific sentence. 2010

Applications of Ampere's law




Ampere's law is used to determine the magnetic field produced by a current distribution with a symmetry, exactly as the Gauss's law is used to determine the electric field produced by a charge distribution with a symmetry. Note that the Gauss's law involves a surface integral of an electric field, whereas Ampere's law involves a line integral to determine a magnetic field.



1. Long straight wire carrying a current I



We want to determine the magnetic field B due to a long straight wire of radius R and current density J; carrying a current I inside the wire (r<R) at the edge of the wire (r = R) and outside the wire (r>R).

The current crossing the wire is I piercing the surface πR2. The current that pierces the surface πr2 is πr2 (I/πR2) = (r2/R2) I.

1.1. r < R:

Ampere's law gives:

∮ B dr = μo (r2/R2) I
∮ B dr = 2πr B. Therefore:
2πr B = μo (r2/R2) I
B = (μoI/2π) (r/R2)

B = (μoI/2πR2) r

1.2. r = R:

Ampere's law gives:

∮ B dr = μo (R2/R2) I
∮ B dr = 2πR B. Therefore:
2πR B = μo I
B = μoI/2πR

B = μoI/2πR

1.3. r > R:

Ampere's law gives:

∮ B dr = μo I
∮ B dr = 2πr B. Therefore:
2πr B = μo I
B = (μo/2π) 1/r

B = (μo/2π) 1/r



1.4. Graph of B(r)



The magnitude of the field increases linearly with distance R from the axis inside the wire. Outside the wire the magnitude of the field decreases inversely with the distance r.



2. Ideal solenoid



A solenoid is formed by winding a long wire onto a cylinder. We are interested to determine the magnetic field produced a solenoid carrying a current, with n the number of turns per unit length (turns/meter).

∮ B dr = ∮(ab) B dr + ∮ B(bc)dr +
(cd)B dr + ∮(da)Bdr = μo Σi

Because dr is perpendicular to B: ∮ B(bc) dr = 0

We can chose (bc) and (da) as long as possible (infinity) and have B there = 0: ∮(cd)Bdr = 0

Because dr is perpendicular to B: ∮(da)Bdr = 0

Only the nonzero part ∮(ab)Bdr contributes to the magnetic field.

∮Bdr = ∮(ab)Bdr = B L

If N is the number of turns within the length L then Σi = NI. N is also equal to nL. Hence Σi = nLI

Ampere's law gives:

∮Bdr = μoΣi, that is B L = μo n L I

We obtain then

B = μonI



Magnetic field of an ideal solenoid currying a current I with n turns per unit length. Magnetic field outside is 0 and inside uniform and parallel to the axis:

B = μonI



3. Force between currents



Consider two parallel long straight wires carrying the currents I1 and I2 separated by a distance R. The current I2 produces the magnetic field B where the first wire is located.

B is tangent to the circle and perpendicular to the segment l that we consider on the first wire. Amper's law gives:

B2 = μoI2/2πR

The magnetic force due to B and I1 is :

F = B2I1l = μoI2 I1l/2πR



Force between currents:

F = μoI2 I1 l/2πR





 


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