Question

μ,NIR 2(R2 +x) 2. (50 pts) The expression for the magnetic field on the axis of a current loop is B,(x) A Helmholtz coil is comprised of two identical coils separated by the distance R, with one coil at the origin and the other shifted over at Xo = R (the radius of the coils), where the current runs counterclockwise in both. It is a useful tool because the magnetic field at the center of the apparatus is large and nearly uniform. (a) Calculate the magnetic field strength B everywhere on the x-axis (expressed in terms of x), then evaluate B at the center of the Helmholtz coil (i.e. at ix R/2). (b) Next, calculate the first derivative of the field (dB,Vdx) and evaluate it at the center of the coils. (c) If, on the other hand, you reverse the current in one of the coils so that it is clockwise compared to the first, you end up with a Marwell coil What is the magnetic field at the center of the pair for a Maxwell coil, and (d) what is the first derivative of the magnetic field (dB,/dx) evaluated at the center? (e) Note any differences in the B fields and their derivatives between the Helmholtz and Maxwell configurations. (f) The quantity dBWdx is useful because it determines the force that is exerted on a magnetic moment u (think button magnet), given by the relationship F pCdB/d). Using the values N= 1000, R = 10 cm, 1-5 A, and μ-0.02 A-m2, what is the actual force on the magnetic moment at the center of the coils in both the Helmholtz and the Maxwell configurations? NOTES: Please express your final answers to parts a, b, c and d in terms of N 1, R and constants only. Before evaluating the expression for B, at x = 12, calculate all derivatives of Br algebraically in terms of x. And finally, all calculations for the Maxwell coil are the same as those of the Helmholtz coil except for one sign difference, so you do not need to do the differentiation a second time.)

Please show all work, thank you.

Answer 1

Magnetic field due to a circular loop radius R carrying current I, at a distance x on the axis of loop is B^- (x) = (/4 pi) (NI 2 pi R^2/(R^2 + x^2)^3/2) x^- one coil is at x = 0 and the other is at x_0 = R B^- (x) = (/4 pi) (2 NI pi R^2) [1/(R^2 + x^2)^3/2 + 1/(R^2 + (x - x_0)^2)^3/2] x^- B^- (x) = (NIR^2/2) [1/(R^2 + x^2)^3/2 + 1/(R^2 + (x - R)^2)^3/2] x^- At center of coil x = R/2, B^- (x) = (NIR^2/2) [2/(5R^2/4)^3/2] x^- dB^- (x)/dx = (NIR^2/2) [(-3/2) (2x/(R^2 + x^2)^5/2) - (3/2) 2 (x - R)/(R^2 + (x - R^2)^5/2)] x^- = (NIR^2/2) [-3 (x - R)/(R^2 + (x - R)^2)^5/2 - 3x/(R^2 + x^2)^5/2] x^- dB^- (x)/dx = (NIR^2/2) [3R/2 - 3R/2/(R^2 + R^2/4)^5/4] x^- at x = R/2 rightarrow Hence at x = R/2 dB^- (x)/dx = 0