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 at x 0. It turns out that d2B/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 x due to one loop, the other loop and the sum of the two, overlayed on a single graph.

Answer 1

2 2 ,R - Ao NIR2 2 ected 2. るト。NTR 2. 2 2

2 NI = :2-9x10 suita 14 2 X 10 Col 2 2