EECE 380 Homework 6 Solution, Fall 2016

Homework 6 Solution Chapter 5 Problem 5.2 Whena particle with charge g and mass m is introduced into a medium with a uniform field B such that the initial velocity of the particle u is perpendicular to B(Fig. P5.2), the magnetic force exerted on the particle causes it to move in a circle of radius a. By equating F,, to the centripetal force on the particle, determine a in terms of g, m, u, and B. © © ®© © ® Figure P5.2: Particle of charge g projected with velocity u into a medium with a uniform field B perpendicular to u (Problem 5.2). Solution: The centripetal force acting on the particle is given by K = mu?/a. Equating F, to Fy given by Eq. (5.4), we have mu? /a = quBsin@. Since the magnetic field is perpendicular to the particle velocity, sin @ = 1. Hence, a = mu/qB. Problem 5.4 The rectangular loop shown in Fig. P5.4 consists of 20 closely wrapped turns and is hinged along the z-axis. The plane of the loop makes an angle of 30° with the y-axis, and the current in the windings is 0.5 A. What is the magnitude of the torque exerted on the loop in the presence of a uniform field B = y2.4 T? When viewed from above, is the expected direction of rotation clockwise or counterclockwise? | SS 7 U2 turns SJ y KS m in > KN ™ Figure P5.4: Hinged rectangular loop of Problem 5.4. Solution: The magnetic torque on a loop is given by T = m x B (Eq. (5.20)), where m= AaN/d (Eq. (5.19)). For this problem, it is given that J = 0.5 A, N = 20 turns, and A=0.2 mx 0.4 m= 0.08 m?. From the figure, fi = —%cos30° + $sin30°. Therefore, m = fi0.8 (A-m?) and T = 0.8 (A-m?) x $2.4 T = —21.66 (N-m). As the torque is negative, the direction of rotation is clockwise, looking from above. Problem 5.12 Two infinitely long, parallel wires are carrying 6-A currents in opposite directions. Determine the magnetic flux density at point P in Fig. P5.12. 1,=6A Ih=6A P —-e 0.5 m - 2m > Figure P5.12: Arrangement for Problem 5.12. Solution: 2 Ho 2 SHo )7¢ 7 Ot2)=9 — (D- 3 Mol 2 Mol B=o7705) 1? amas 5.15 A circular loop of radius a carrying current /, is located in the x—y plane as shown in Fig. P5.15. In addition, an infinitely long wire carrying current J; in a direction parallel with the z-axis is located at y = yo. N N parallel to z P= (0,0, h) x I, Figure P5.15 Problem 5.15. (a) Determine H at P = (0,0,/). (b) Evaluate H fora =3cm, yo = 10cm, h=4cm, J, = 10 A, and h = 20 A. Solution: (a) The magnetic field at P = (0,0,/) is composed of H; due to the loop and H> due to the wire: H=H,+H. Eq. (5.34) applies to a current in CCW direction, but in the loop of Fig. P5.15, the current is CW. Hence, with z = /) and adding a minus sign, we have la’ H, =2——"“ __ 14724 RpP (A/m). From (5.30), the field due to the wire at a distance r = yo is h I= 2myo where 6 is defined with respect to the coordinate system of the wire. Point P is located at an angel @ = —90° with respect to the wire coordinates. From Table 3-2, $ = —Xsing + feos = (at @ = —90°). Hence, —la* . b Z x A/m). 2(a2 + h?)3/2 + 2m (A/m) (b) H = —236+X31.83 (A/m). Problem 5.21 Current J flows along the positive z-direction in the inner conductor of a long coaxial cable and returns through the outer conductor. The inner conductor has radius a, and the inner and outer radii of the outer conductor are b and c, respectively. (a) Determine the magnetic field in each of the following regions: 0 <r <a, a<r<b, b<r<c,andr>c. (b) Plot the magnitude of H as a function of r over the range from r = 0 to r= 10cm, given that 7 = 10 A,a=2 cm, b=4 cm, andc =5 cm. Solution: (a) Following the solution to Example 5-5, the magnetic field in the region r < a, 2 rT H= ; 6 2na? and in the region a <r <b, H-$— ~ "Onr’ The total area of the outer conductor is A = 2(c* — b”) and the fraction of the area of the outer conductor enclosed by a circular contour centered at r = 0 in the region b<r<cis n(?—b?) rr —b m(c2—b?) c2—b? The total current enclosed by a contour of radius r is therefore > rb ear Tenclosed = I (1 _ ) =I Cb cb” and the resulting magnetic field is + Tenclosed » I ear H= = . ° 2ar oor ( - =) For r > c, the total enclosed current is zero: the total current flowing on the inner conductor is equal to the total current flowing on the outer conductor, but they are flowing in opposite directions. Therefore, H = 0. (b) See Fig. PS.21. Magnetic field magnitude H (A/cm) 0.8 0.7 0.6 0.5 04 Radial distance r (cm) Figure P5.21: Problem 5.21. Problem 5.28 A uniform current density given by J=ih (Alm?) gives rise to a vector magnetic potential A=-z Hod (42 +3?) (Wb/m) (a) Apply the vector Poisson’s equation to confirm the above statement. (b) Use the expression for A to find H. (c) Use the expression for J in conjunction with Ampére’s law to find H. Compare your result with that obtained in part (b). Solution: (a) 5 2 2 PP J VAS SVE +902, +202 <2 ( )| ° _ 2,2 oe oy? +32 Hoy (ty | ~ J . = two? (242) = —ZUoJo. Hence, V7A = —[oJ is verified. (b) 1 1 [. (0A: OA, . (OAy OA: . (OA, AAx = Vv = = _ yy Be A io (SS B30 (3 me (S ay _ 1 (gad: _ 404: ~~ Uo oy vox a1 fg? (pp 2-4yy) 92 (p19? 2 492 ~ ho Ik ( Mo ( +¥") } —¥5_ | Hog @ +y¥) (©) pH-dl=I= [o. $ Ho 420 = Jy nr, H=$H, =bh5. uy Figure P5.28: Current cylinder of Problem 5.28. We need to convert the expression from cylindrical to Cartesian coordinates. From Table 3-2, : .. . . , x @ = —Xsing + ycosd = -kK—=—— + J ——_, Xx? +? x? +32 r=Vx+y. Hence 7 y . x J 5 VJ XJ H= | -%—=—= + y= ]- 9 ety as 98, which is identical with the result of part (b). Problem 5.30 In the model of the hydrogen atom proposed by Bohr in 1913, the electron moves around the nucleus at a speed of 2 x 10° m/s in a circular orbit of radius 5 x 10~!! m. What is the magnitude of the magnetic moment generated by the electron’s motion? Solution: From Eq. (5.69), the magnitude of the orbital magnetic moment of an electron is |mo| = |—4eur| =4 x 1.6« 10°? x 2x 10°x 5x10 =8x107 (Am’).