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NOTES AND DISCUSSIONS The magnetic field produced at a focus of a current-carrying conductor in the shape of a conic section C. Christodoulidesa兲 Department of Physics, School of Mathematical and Physical Sciences, National Technical University of Athens, Zografou, GR 157 80 Athens, Greece 共Received 21 June 2009; accepted 30 June 2009兲 We determine the magnetic field at the focus of a conic section due to a current along the conic section. For a given current I, it is shown that this magnetic field is the same and equal to 0I / 2p for all conics with the same semilatus rectum p. © 2009 American Association of Physics Teachers. 关DOI: 10.1119/1.3183888兴 I. INTRODUCTION The calculation of magnetic fields in textbooks using the Biot–Savart law is limited mainly to current-currying conductors consisting of linear or circular sections. In rare cases the field at the focus of a current-carrying elliptical or parabolic conductor is calculated.1 Recently the field of a more general plane current loop was considered.2 The purpose of this paper is to demonstrate an interesting common property of conic sections regarding the magnetic field produced at one of the foci of a current-carrying conic section. II. ANALYSIS We consider a current-carrying conductor in the shape of a conic section 共a circle, ellipse, parabola, or hyperbola兲. Polar coordinates 共r , 兲 are used to describe the curves, with a focus of the conic section taken to be the origin O, and the axis of symmetry of the curve coinciding with the polar axis 共see Fig. 1兲. There is a constant current I in the conductor in the direction of increasing . A current element Idr with position vector r relative to O produces a magnetic field dB at O, which is given by the Biot–Savart law as dB = 0I r ⫻ dr . 4 r3 共1兲 dr̂ = e d 共4兲 is the unit vector in the direction of increasing . Therefore r̂ ⫻ p dr = r̂ ⫻ e d 1 + e cos 共5兲 r̂ ⫻ p dr = ẑ, d 1 + e cos 共6兲 or where ẑ = r̂ ⫻ e is the unit vector normal to both r̂ and e, and is out of the plane of the curve in Fig. 1. Equations 共2兲 and 共6兲 and the definition r = rr̂ give r⫻ 冉 p dr = d 1 + e cos 冊 2 ẑ 共7兲 so that the ratio in Eq. 共1兲 becomes 冉 冊 1 e r ⫻ dr + cos dẑ. = r3 p p 共8兲 The total magnetic field at O for a circle, ellipse, or parabola is found by integrating Eq. 共1兲 from = − to For the geometry and current in Fig. 1, the direction of dB is normal to the plane of the figure and outward. In polar coordinates the general equation of conic sections 共position vector of a point on the curve兲 is3 r= p r̂, 1 + e cos 共2兲 where r̂ is a unit vector from O to the current element, the length p is the semilatus rectum of the conic, and e is its eccentricity 共e = 0 for a circle, 0 ⬍ e ⬍ 1 for an ellipse, e = 1 for a parabola, and e ⬎ 1 for the left branch of a hyperbola兲. If we differentiate r with respect to , we obtain pe sin dr̂ p dr = . 2 r̂ + d 共1 + e cos 兲 1 + e cos d 共3兲 Fig. 1. A current element Idr on a conic section and the magnetic field dB it produces at O, the focus of the curve. The quantity 1195 Am. J. Phys. 77 共12兲, December 2009 http://aapt.org/ajp © 2009 American Association of Physics Teachers 1195 Fig. 2. The two branches of a hyperbola. B= 0I ẑ 4 冕冉 − 冊 1 e + cos d . p p Fig. 3. Conic sections with a common focus O and the same value of the latus rectum 2p. 共9兲 If instead of the equation of a conic such as that in Eq. 共2兲, we had used the general form r = r共兲r̂ 共10兲 for a curve in plane polar coordinates, Eq. 共9兲 would have the more general form B= 0I ẑ 4 冖 d , r r= B− = − 2 derived in Ref. 2. We perform the integration in Eq. 共9兲 and obtain 共12兲 for a circle, ellipse, or parabola. The hyperbola has to be treated separately because it has two branches between its two asymptotes, which form angles of 0 and − 0 with its axis of symmetry 共see Fig. 2兲. The value of 0 is given by cos 0 = −1 / e because this value of cos makes r infinite. The field produced by the left branch of the hyperbola is B+ = 2 0I ẑ 4 冕冉 0 0 冊 1 e 0I + cos d = 共0 + e sin 0兲ẑ. p p 2 p 共13兲 Equation 共2兲 produces negative values of r for 0 ⱕ ⬍ 2 − 0. We can keep r positive if we change the sign of the right-hand side of Eq. 共2兲 and can limit to the range −共 − 0兲 ⱕ ⱕ − 0 for the right branch of the hyperbola if we write cos共 − 兲 instead of cos in Eq. 共2兲. The right branch of the hyperbola is then given by 1196 Am. J. Phys., Vol. 77, No. 12, December 2009 共14兲 for −共 − 0兲 ⱕ ⱕ − 0. For the current direction shown in Fig. 2, the magnetic field produced at O by the right branch of the hyperbola is 共11兲 0I ẑ B= 2p p r̂ e cos − 1 = 0I ẑ 4 冕 冉 −0 0 − 冊 1 e + cos d p p 0I 共 − 0 − e sin 0兲ẑ. 2 p 共15兲 The total magnetic field produced by a hyperbolic conductor at its focus is therefore B = B+ + B− = 0I ẑ, 2p 共16兲 the same as for the other conic sections. The field at the right-hand focus of the hyperbola is also given by Eq. 共16兲. We conclude that conic sections with the same semilatus rectum p, carrying the same current I, produce the same magnetic field 0I / 2p at their foci. A group of such conic sections is shown in Fig. 3 with their common focus at O. a兲 Electronic mail: cchrist@central.ntua.gr J. D. Kraus, Electromagnetics, 4th ed. 共McGraw-Hill, New York, 1991兲, Problems 6-3-10 and 6-3-12. 2 J. A. Miranda, “Magnetic field calculation for arbitrarily shaped planar wires,” Am. J. Phys. 68共3兲, 254–258 共2000兲. 3 M. R. Spiegel, Mathematical Handbook of Formulas and Tables 共McGraw-Hill, New York, 1968兲, pp. 37–39. 1 Notes and Discussions 1196