Ampere Balance

The Ampere Balance

The Ampere Balance is made up from 2 parallel horizontal wires. Wire T is above wire H. Wire T makes a balanced pendulum The upper wire segment is part of a physical pendulum that can be balanced at a specific distance from the lower wire. A magnet N is used to damp the oscillations of the pendulum (using induced currents). On the tray T one places a mass m. Then a current I goes through the two wires T and H  as to restore the wire spacing to the initial value of d (when the magnetic force equals the weight mg).

Questions

a) Please find the direction of the current in the upper wire for repulsive interaction between the wires of the Ampere balance. Draw a diagram.

b) Please tell why there is no force on wires N and G due to segment H.

c) Find the magnetic force on wire T, that depends on d, L and current I.

d) For values of d = 4 mm and L = 26 cm, please find the value of the current to balance a mass m=2 mg.

You can find an exact description of this balance here.

ampere balance

Principle

a)

By DEFINITION two parallel wires that have PARALLEL currents attract each other. If the currents are ATNTI-PARALLEL the two wires repel each other. (This is the principle on which the Ampere balance works.)

The lower wire creates a field B according to the rotation of a right screw (screw is advancing in the direction of the current and its rotation gives the direction of B).

The upper wire feels a magnetic force $F = L*(I times B)$, where $times$ is the vector product. Hence the magnetic force is attractive.

Perpendicular wires

b)

Still in the wires N and G the currents are perpendicular to the field $B$ (so the vector product $I times B$ should be different from zero), but in this case the currents in the wires N and G are opposing one to the other so in total the two magnetic forces on the wires N and G will cancel each other.

Magnetic force

c)

The magnetic field of a current $I1$ through a wire is given at distance $d$ from the wire by the equatiion:

$B = mu*I1/(2*pi*d)$

Therefore, the magnetic force of field $B$ on current $I2$ is

$F =| L (I2 times B) |= B*I2*L$

because the angle between $B$ and $I$ is 90 degree.

Hence $F = mu I1*I2*L/(2 pi d) = mu I^2 L/(2 pi d)$

where $mu$ is the magnetic permeability of the air $mu=4*pi*10^{-7} H/m$

d)

The condition of equlibrium is weight=magnetic force.  Because of this:

$m*g = mu I^2L/(2pi d)$

$I = sqrt{(2pi d mg)/mu*L} =sqrt {(0.004*2*10^{-6}*9.81) / 4*pi*10^{-7}*0.26/ (2*pi)} = 1.2285 A$