Question

The magnitude of the force on a magnetic dipole μ aligned with a non uniform magnetic field in the positive x-direction is Fx- Suppose two flat loops of wire each have radius R and carry a current dx (a) The loops are parallel to each other and share the same axis. They are separated by a variable distance x » R. Show that the magnetic force between them varies as (Submit a file with a maximum size of 1 MB.) Choase File No file chosen This answer has not been graded yet. (b) Find the magnitude of this force in N), taking I10.5 A, R0.520 cm, and x4.50 cm

Answer 1

B x Given by B x = mu_0/4 pi times 2 pi R^2 I/(x^2 + R^2)^3/2 x^2 + R62 almostequalto x^2 [x > > R] B x = mu_0 I R^2/2 x^3 f = Vertical-bar mu Vertical-bar d B x/dx d B/dx = -3 mu_0 I R^2/x^4 [(x^-3)^� = -3 x^-4] Vertical-bar f Vertical-bar = I times pi R^2 times 3 mu_0 I R^2/2 x^4 = x/x^4 Vertical-bar f Vertical-bar alpha 1/x^4 I = 10.5 A, R = 0.520 cm x = 4.50 cm Vertical-bar f Vertical-bar = 3/2 I^2 R^4 mu_0 pi/x^4 mu_0 = 4 pi times 10^-7 = 3/2 (10.5)^2 (0.0052 m)^4 times 4 pi times 10^-7 times pi/(0.045)^4 = 11641010.338 N = 0.1164 times 10^-8 N Vertical-bar f Vertical-bar = 0.1164 times 10^-8 N