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Problem 3 a) Find the magnetic field H at the center of a square loop carrying a current I. The side of the square loop is b meters long. b) A circular loop that has a radius a and that carries a current I produces the same magnetic field strength at its center as that at the center of a square loop that has side b and that carries the same current I. Determine the ratio of b to a. c) The Earth's magnetic field at the north magnetic pole is approximately 0.62 G (1 G = 104 Wb/m2). Assume that this magnetic field is produced by a loop of current flowing along the equator. Estimate the magnitude of this current. The radius of the Earth is approximately 6,500 km.

Answer 1

Magnetic field at center of square Loop Magnetic field due to straight current carrying conductor at point P B = mu_0 I/4 pi b (sin alpha + sin beta) hat a_phi hat a_phi direction of Magnetic field. hat a_phi = [unit vector along current direction] * [unit vector 1^lar direction] down arrow cross product b = distance from conductor to point p . = mu_0 H Therefore H = B/mu_0 rightarrow mu_0 I/4 pi b [sin d + sin beta] hat a_phi/mu_0 Therefore H for point at P = I/4 pi b (sin d + sin beta) hat a_phi Total . H at center P = H_1 + H_2 + H_3 + H_4 H_1 rightarrow H from conductor 1 to center of square (p). The distance is same from all the conductors (b/2) direction. \r\n Therefore H^-_T = 4 H_1 H_1 = I/4 pi (b/2) [sin (alpha) + sin bet] hat a_