## Loop Shunts - design information

One of the most common schemes for making a resistive shunt is the loop shunt.
The idea is that by folding the resistive element so that the opposite currents
are next to each other, their magnetic fields cancel, reducing the inductance.
However, the forces on the shunt can be significant. The following is a drawing
of such a shunt, with current I flowing from left to right.

#### What's the resistance of this shunt, and how much power will be dissipated?

If we assume that d is 1 cm (about 1/2 inch), and L is 10 cm (4"), and
we're using standard 1x1/8" bar stock, the resistance will be:

R = 1.71E-6 ohm cm * 21 cm/0.8 cm^2 = 0.045 mOhm.

At a current of 10 kA, the heat dissipated will be about 4.5 kW. A 10 microsecond
pulse at that current will put about 0.045 Joules into the shunt, which is negligible.

#### What's the force F tending to pull the shunt apart?

Using Ampere's law, the force between two parallel conductors is: F = I*L/d,
where F is in Newtons, I is in Amps, and Land d are in Meters (that is, 1 amp
creates a force of 1 N between two wires 1 meter long). Plugging in some reasonable
numbers, and assuming that the current carrying member is infinitely thin:

I = 10 kA, d = 0.01 m, L = 0.1 cm

F = 10000 * 0.1/0.01 = 100 kN, which is a substantial force (about 22,000 pounds).

If the shunt were made from copper bar that is 1"x1/8" (a typical
size), let's figure out the shear load at point "P", assuming that
the bar is perfectly rigid and doesn't flex to absorb some of the force. The
cross sectional area will be 1/8 square inch, or about 0.8 sq cm, so the load
is 100 kN/8E-5 = 1.24E9 N/sq meter (Pascal). The yield strength of hard copper
is about 231 MPa (with ultimate strength of 351 MPa), although it varies somewhat
with cold working, etc. (from http://www.matweb.com/
) In any case, the force on the joint is more sufficient to break the copper,
or, at the very least, to permanently deform the members.

loopshunt.htm / 26 Sep 2000 / Back
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Jim