Question

Helmholtz coils are a method developed to provide a uniform magnetic field in a small region. They are constructed by placing two identical wire loops parallel to each other, separated by a distance equal to their radii, as shown in the Figure. Assume that the two loops are also connected in parallel to a power supply, so that they produce the same field.

(a) Find the length of a 28-gauge copper wire that will produce a 75 Ω coil.
m
[Use the American Wire Gauge information on Wikipedia to help solve this.]

If each coil has a 5 cm radius, then how many full turns of wire must you wind onto the spool to produce this resistance?


(c) If your power supply has a maximum output of 30 V, then what is the maximum current that will flow through the coils?
  A

(d) What is the magnetic dipole moment of one of the coils?
  A·m²

(e) What is the magnetic field at the midpoint of the line that connects the centers of the coils?

Answer 1

(a) The resistance per unit length of a 28 Gauge Copper wire(AWG 28) is 212.9m/m=212.9*10^-3/m

Thus, to produce a 75 coil, the length of the copper wire required=l=75/212.9*10^-3=352.28 m

(b) Radius of each coil(r)=5 cm=0.05 m

Thus, the perimeter of the loop L'=2r=0.314 m

Thus, no of turns of wire to wind up onto the spool to produce the resistance(N) =352.28/0.314=1121.91=1192(nearly)

(c) Maximum output of Power supply V=30V

The maximum current flowing through the coils I=V/R=30/75=0.4A

(d) The magnetic dipole moment of the coils m=IA

where A=area of the coil=r²=3.14*0.05²=0.00785 m²

Thus, m=0.4*0.00785=0.00314 Am²

(e) Magnetic field at the midpoint connecting centre of the coils B=8oI/(11.18 r)

where o= permeability of free space=4*10^-7N/A²

Thus, B=(8*4*10^-7*0.4)/(11.18*0.05)=7.19 *10^-6T