Question

A current I flows around a circuit in the shape of a regular hexagon of side s (a) Determine the magnetic field at the center of the hexagoin (b) Evaluate for the case 1 12 A., s -15 cm. (c) Compare your result with the magnetic field created at the center of a circle of radius s, and comment.

Answer 1

Magnetic field due to a finite length of a wire is given by mu_0 I/2 pi r (sin theta_1 + sin theta_2) Her theta_1 = theta_2 = pi/n where n is number of sides Hexagonal = 6 theta_1 = theta_2 = pi/6 = 30 degree tan theta_1 = r/S/2 tan 30 degree = r/S/2 r = 2/squareroot 3 S So B_1 = mu_0 I/4 pi (2S/squareroot 3) (sin 30 degree + sin 30 degree) B_1 = squareroot 3 mu_0 I/4 pi (2S) (1/2 + 1/2) = squareroot mu_0 I/8 pi S Due to 6 line segment field is equal so B = 6B_1 = 3 squareroot 3 u_0 I/4 pi S = 3 squareroot 3 u_0 I/4 pi S \r\n for J = 12 A S = 15 cm B = 3 squareroot 3 mu_0 I/4 pi S mu_0/4 pi = 10^-7 B = 3 squareroot 3 times 10^-7 times 12/1S times 10^-2 B_Hexa = 4.16 times 10^-5 Tesla (c) Magnetic field at centre of radius B_circle = mu_0 I/2 pi S = 2 times 10^-7 times 12/1S times 10^-2 = 1.6 times 10^-5 Tesla B_Hexa > B_circle