Question

4) With FlipltPhysics you have already found the field due to a circular loop of radius R carrying current I at a point a distance x from the center of the loop but along the axis. A pair of such loops placed a distance R apart makes up a Helmholtz coil. Such a coil is used in the e/m ratio lab (a) Determine the magnetic field as a function of x along the axis of a Helmholtz coil, with x = 0 located in the middle between two coils (b) Show that dB/dx 0 atx-0. It turns out that d2 B/dx2 is also zero, but I will spare you the tedious calculation. These derivatives show that the field is fairly constant near the center (c) In the lab write-up, you are given the magic formula of B = 4.23 × 10-31. Reverse engineer the number of turns in each coil, assuming they are identical. The diameter of the coil is 15.5 cm (d) Make a graph of the magnetic field as a function of χ due to one loop, the other loop and the sum of the two, overlayed on a single graph.

Answer 1

4. netic -field (nt- a 2 opace a) 040 1 col o Ceil tte ts csi ls tha maz nehe freed due The mhehe eld 2

Mon IR 5R1 8Mon TR1 3 52 R

(d)

0.8 0.6 Left Coil Right Coil _ Both Coil 0.4 0.2 4 10

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