# 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.

#### 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:

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

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

Mehdi Mehdizade

August 26, 2017 @ 1:54 pm

At the last line, it must be I eaquals (2pdmg/uL), so the answer is 1/0.814, I equal to 1.2285 A

valentin68

August 26, 2017 @ 8:56 pm

You are absolutely right! We have now corrected it. Thank you for that!