Chapter 28--ExamplesProblemParallel DirectionsOpposite Directions DirectionsSlide 5Biot-SavartDirectionSlide 8RH RuleThe MagnitudeSlide 11First, r<=aNext, a<=r<bb<=r<=cr>c1Chapter 28--Examples2ProblemTwo long, parallel transmission lines, 40 cm apart, carry 25 A and 75 A currents. Find all locations where the net magnetic field of these wires is zero if these currents are a) in the same direction and b) in opposite directions3Parallel DirectionsBased on the figure, the only place that the fields can be zero is between wires.Assume that the zero point is a distance x from the 75 A wire. So now set B75=B25O O OX X XX X XX X XO O OO O OX X X75Ax mxxxxxIxIrIB3.01003025)4.0(*754.022225075004Opposite Directions DirectionsBased on the figure, there are two regions outside the wires.Since one current is smaller than the other, we must choose the region closest to the smaller current.Assume that the zero point is a distance x from the 25 A wire. So now set B75=B25O O OX X XX X XX X XO O OX X XX X X75Ax mxxxxxIxIrIB2.0501075)4.0(*254.0222750250025A5ProblemConsider the figure below. The curved segments are arcs of circles of radii a and b. The straight segments are along radii. Find the magnetic field at point P (the center of the curvature) assuming i as the current.abii6Biot-Savart2ˆ4 rrsdiBdoabiiBased on Biot-Savart, the current segments along the radial vector do not contribute since they are parallel with r.Now we must integrate across each arc. The integration variable is .ds=rd biBbrataiBaratrirdiBrrdirrsdiBdoboaoooo44444ˆ400227DirectionabiiBy the RH Rule, the current from b is out of the page at Point P.The current from a is into the page.We will choose out of the page as positive and the magnetic field at P isabiBaibiBBBoooab114448ProblemFour long copper wires are parallel to each other, their cross-sections forming the corners of a square with sides a=20 cm. A 20 A current exists in each wire in the direction shown in the figure below. What are the magnitude and direction of B at the center of the square. XaaaaX9RH RuleXaaaaXThe lower right and upper left corners each point to the upper right.The fields from the lower left and upper right point to the upper left.Thus the total field point up the page10The MagnitudeThe distance to the center from any corner is a/sqrt(2). All of the horizontal components have added out. Only the four vertical components sum toTBaiaiaiBooo570108)2.0()20)(104(22222*445cos22*411ProblemA solid conductor with radius a is supported by insulating material on the axis of a conducting tube with inner radius b and outer radius c (see figure below). The central conductor and the outer conductor carry an equal current i in opposite directions. However, the magnitude of current density in the inner conductor is expressed by J=kr where k is an appropriate constant. Derive an expression for the magnetic field at all points (r<a, a<r<b, b<r<c, r>c)Derive an expression for current density in the outer conductor.Show that each boundary is matched.12First, r<=aAt r=a, total current is i so i=J*AJ=ka and A=a2 so i=ka3 or k=i/a3 Using RH rule, with our thumb pointing in direction of current, our fingers are matching direction of integration so this is a positive direction  aiaaiBaratariBarirBarirairkrJAirBsdBisdBenclosedenclosed2222*2032032033033332013Next, a<=r<b biBaiBbrataratriBiirBsdBisdBenclosedenclosed2222000014b<=r<=cJ’=-i/AA=(c2)-(b2)J’=-i/[(c2)-(b2)]                        022222222002222022220bcbciicBbratbibbbciibBbratbrbciirB                      2222022220002202'2''2brbciirBbrbciiiirBbrAAJiiiirBsdBisdBenclosedenclosed15r>c 0020BiiirBsdBisdBenclos edenclosedRecall for a coaxial cable with equal static charges, E=0 outside as