Homework 6 solutions PHYS 212 Dr. Amir


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1 Homework 6 solutions PHYS 1 Dr. Amir Chapter (II) A rectangular loop of wire is placed next to a straight wire, as shown in Fig There is a current of.5 A in both wires. Determine the magnitude and direction of the net force on the loop. The magnetic field at the loop due to the long wire is into the page, and can be calculated by Eq The force on the segment of the loop closest to the wire is towards the wire, since the currents are in the same direction. The force on the segment of the loop farthest from the wire is away from the wire, since the currents are in the opposite direction. Because the magnetic field varies with distance, it is more difficult to calculate the total force on the left and right segments of the loop. Using the right hand rule, the force on each small piece of the left segment of wire is to the left, and the force on each small piece of the right segment of wire is to the right. If left and right small pieces are chosen that are equidistant from the long wire, the net force on those two small pieces is zero. Thus the total force on the left and right segments of wire is zero, and so only the parallel segments need to be considered in the calculation. Use Eq. 8. I I I I F F F l l I I l net near far near far 1 d d d d near far near far T m A A m N, towards wire 0.00 m m 1. (II) A coaxial cable consists of a solid inner conductor of radius R, 1 surrounded by a concentric cylindrical tube of inner radius R and outer radius R (Fig. 8 4). The conductors carry equal and opposite currents I 0 distributed uniformly across their cross sections. Determine the magnetic field at a distance R from the axis for: (a) R < R ; (b) R < R < R ; (c) R < R < R ; (d) 1 1 R > R.
2 Homework 6 solutions PHYS 1 Dr. Amir (e) Let I A, R cm,.00 cm, R.00 cm. R and R.50 cm. Graph B from R 0 to R 1 r R I 0 (out) R I 0 (in) Because of the cylindrical symmetry, the magnetic fields will be circular. In each case, we can determine the magnetic field using Ampere s law with concentric loops. The current densities in the wires are given by the total current divided by the crosssectional area. I0 I0 Jinner J outer R R R 1 (a) Inside the inner wire the enclosed current is determined by the current density of the inner wire. B ds 0Iencl 0 J inner R I R I R BR B R1 R1 (b) Between the wires the current enclosed is the current on the inner wire. 0I0 B ds 0Iencl B R 0I0 B R (c) Inside the outer wire the current enclosed is the current from the inner wire and a portion of the current from the outer wire. B ds 0Iencl 0 I0 J outer R R R R R R Br I I B 0I R R R R R (d) Outside the outer wire the net current enclosed is zero. B ds 0Iencl 0 BR 0 B 0
3 Homework 6 solutions PHYS 1 Dr. Amir (e) See the adjacent graph B (105 T) (II) Consider a straight section of wire of length d, as in Fig. 8 48, which carries a current I. (a) Show that the magnetic field at a point P a distance R from the wire along its perpendicular bisector is R (cm) B = μ 0I d πr (d +4R ) 1 (b) Show that this is consistent with Example 8 11 for an infinite wire. (a) Choose the y axis along the wire and the x axis passing from the center of the wire through the point P. With this definition we calculate the magnetic field at P by integrating Eq. 85 over the length of the wire. The origin is at the center of the wire. 1 d ˆ ˆ ˆ 0I d l rˆ 0I d l r dy R y 0I j i j B / 4 r 4 r 4 1 d R y 0IR kˆ 4 1 d 1 d R dy y / d / IR ˆ y I d k 4 R R y 4R d 0 0 1/ 1/ R d / (b) If we take the limit as d, this equation reduces to Eq kˆ
4 Homework 6 solutions PHYS 1 Dr. Amir 0I d 0I B lim d 1/ R 4R d R Chapter 9. (I) The rectangular loop shown in Fig. 9 7 is pushed into the magnetic field which points inward. In what direction is the induced current? As the coil is pushed into the field, the magnetic flux through the coil increases into the page. To oppose this increase, the flux produced by the induced current must be out of the page, so the induced current is counterclockwise. 0. (II) If the U shaped conductor in Fig. 9 1a has resistivity whereas that of the moving rod is negligible, derive a formula for the current I as a function of time. Assume the rod starts at the bottom of the U at t 0, and moves with uniform speed v in the magnetic field B. The crosssectional area of the rod and all parts of the U is A. The emf is given by Eq. 9 as e Bv l. The resistance of the conductor is given by Eq. 5. The length in Eq. 5 is the length of resistive material. Since the movable rod starts at the bottom of the U at time t = 0, in a time t it will have moved a distance vt. I v Blv Blv BlvA e Bl v Bl v Bl va R L A A vt l vt l 48. (II) A modeltrain transformer plugs into 10V ac and draws 0.5 A while supplying 7.5 A to the train. (a) What voltage is present across the tracks? (b) Is the transformer stepup or stepdown?
5 Homework 6 solutions PHYS 1 Dr. Amir (a) Use Eqs. 95 and 96 to relate the voltage and current ratios. V N I S S S N V P S I I 0.5A P P ; V V S P 10V 5.6V V N I N V I I 7.5A P P P S P S S (b) Because V V, this is a stepdown transformer. S P
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