# Exam II. Solutions. Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points. The wires repel each other.

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1 Exam II Solutions Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points. 1.! Concerning electric and magnetic fields, which of the following is wrong?!! A time-independent B-field does no work on a single moving charge.! E-field lines never form closed curves.! [Lines of induced E-fields form closed curves.]! An E-field changing with time creates a B-field.! A B-field changing with time creates an E-field. 2.! Two parallel straight wires carry equal currents directed as shown.!! The B-field at a point halfway between them is zero. [It is a maximum there.]! The B-field at a point just below the bottom wire is directed into the page. [Out.]! The wires repel each other.! None of the above is true.

2 3.! A copper plate rests on a frictionless table; a bar magnet a small distance above it moves to the right, as shown from above. In case A the magnet moves into the space directly above the plate; in case B it moves away from that space.! The plate accelerates to the right in both cases.! The plate accelerates to the left in both cases.! The plate accelerates to the right in case A and to the left in case B.! The plate accelerates to the left in case A and to the right in case B. [The forces oppose relative motion of magnet and conductor.] 4.! A small loop of wire is at rest in a non-uniform B-field, with the plane of the loop perpendicular to the field. The B-field strength is increasing with time. Which of the following is wrong?! The induced magnetic moment of the loop is opposite to the given field.! The force on the loop is toward the region of stronger field. [Weaker.]! The loop moves in such a way as to decrease the potential energy.! One of the above is not true. Part B. True-false questions. Check T or F depending on whether the statement is true or false. Each question carries a value of 3 points. 5.! In the situation shown, if V b V a = 12 V, the battery is supplying energy to the rest of the circuit. T F a b 10 V 1 Ω [Since the potential rises through the resistor, the current is running from b to a.] 6.! The directions specified by Lenz s law are required by conservation of energy. T F 7.! If I is decreasing in the situation shown, then V a > V b. T F [The inductor is putting energy into the circuit.] I a b

3 Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method clearly. Problems carry the point values shown. 1 In the circuit shown, find the terminal voltage V b V a : Before the switch is closed. After the switch is closed.! [10 points] 12 V 2 Ω a b 6 Ω 3 Ω The current through the battery is 12/8 = 1.5 A, so the terminal voltage is 12 3 = 9 V. The effective resistance of he two in parallel is 1/R eff = 1/6 + 1/3 = 1/2, or R eff = 2 Ω. The current through the battery is 12/4 = 3 A, and the terminal voltage is 12 6 = 6 V. 1 A particle of positive charge is moving up the page with velocity v as shown. It is in a region of uniform magnetic field into the page, indicated by the shading.! [10 points] Sketch the circle the particle follows and show that the magnetic moment of the loop represented by this motion is opposite to the B-field. Suppose the particle has negative charge. Sketch its path also, and show that the magnetic moment in that case is still opposite to the B-field. B v (a) (b) The direction of v B is to the left. For a positive charge, this means the force qv B is to the left, and it will run in the counter-clockwise circle shown. This current loop has a magnetic moment out of the page, opposite to B. Now the force is to the right, so the particle runs in the clockwise circle shown. But because the charge is negative, this represents a counter-clockwise current loop, so the magnetic moment is again out of the page, opposite to B.

4 2.! If v << c one can adapt the Biot-Savart law to calculate the B-field created by a moving point charge, replacing Idl by qv so that the field at displacement r from the moving charge is given by B = µ 0 qv r 4π r 3. Use this formula and consider the two positive charges, shown at a particular instant. Find the direction of the B-field created by q 1 at the location of q 2, and the direction of the magnetic force q 1 exerts on q 2. q 1 v 1 v 2 r q 2 Find the direction of the B-field is created by q 2 at the location of q 1, and the direction of the magnetic force q 2 exerts on q 1. [15 points]! What law of Newtonian mechanics does this situation seem to violate? {You are not asked to give the resolution of the problem.] In this case r (directed from q 1 to q 2 ) is parallel to v 1, so B(at q 2 ) = 0. There is no magnetic force on q 2 due to q 1. Now r is directed from q 2 to q 1, so B(at q 1 ) is out of the page. The force q 1 v 1 B is down the page. Newton s 3rd law is violated, because the two forces are not equal and opposite.! [The importance of this violation of the 3rd law is that the total momentum of the system of the two charges is not conserved, despite there being no outside forces. What is missing is the momentum contained in the electromagnetic field. When that is included as part of the system, the total momentum is in fact conserved.]

5 3.! In the primitive motor situation shown, the massless conducting horizontal bar can slide on frictionless vertical conducting rails. The battery emf and the B-field are fixed but the resistance R can be varied. Give answers in terms of M, g, B, E and l. Call R 0 the resistance for which the bar and weight remain at rest. Find R 0. [What is the current?} R B l M E How much power is delivered by the battery, and where does it go? The resistance is gradually reduced to R 1, and the weight now moves upward with constant speed v. Find R 1. [What is the emf E produced by the upward motion of the bar?] d.! Show that the total power delivered by the battery is the same as in (b). Explain the difference in where it goes.! [20 points] The upward force is that exerted by the B-field on the current carrying bar: F m = I l B, with magnitude IlB. This must balance the weight, so IlB = Mg, or I = Mg/lB. The loop rule gives E IR 0 = 0, so R 0 = E /I = ElB/Mg. The power from the battery is EI = EMg/lB. It goes entirely to the resistor in Joule heating. [One can verify that it is the same as I 2 R 0.] Now the rising bar produces an induced emf E = Blv. This opposes the current. If the weight moves up with constant speed, the total force on it is still zero, so the current must be the same as before. The loop rule now gives E E IR 1 = 0, or R 1 = (E Blv)/I = (Bl/Mg)(E Blv). [This is the same as R 1 = R 0 (B 2 l 2 /Mg) v.] d.! Since the current and the emf are the same, the power EI from the battery is the same. Now only part of it (I 2 R 1 ) goes to Joule heating, while the rest goes to raising the weight. [One can verify that EI I 2 R 1 = Mgv, which is the power to raise the weight.]! [For any reasonable numbers, the fraction of the power from the battery that goes to lifting the weight is tiny. This is a very bad design for a motor.]

6 4.! Shown is a cross section of a coaxial cable with a hollow inner conductor of radius a and an outer conducting sheath of radius b. The two conductors carry current I in opposite directions. Use Ampere s law to find the B-field magnitude as a function of r, the distance from the axis of the cable, in three regions: (1) within the inner conductor; (2) in the gap between the conductors: (3) outside the cable. [You can assume the inner conductor carries current into the page.] What is the magnetic energy density in each region, as a function of r? Find the magnetic energy stored in length l of the cable. [Use as volume element a cylindrical shell of radius r, thickness dr, and length l.] d.! Use the formula for stored magnetic energy to find the self-inductance in length l of the cable.! [20 points] By the symmetry, the B-field lines will be circles around the axis of the cable. Choose a circular path of radius r. Then! B dr =! Bdr = B! dr = B 2πr. In (1) there is no current at all, so I linked = 0, and we have B = 0. In (3) the path links both conductors, so the net I linked = 0 is zero and we again find B = 0. In (2) the path links only the inner conductor, so I linked = I and we find B(r) = µ 0 I/2πr. In (1) and (3) the energy density is zero. In (2) we have u m = B2 2µ 0 = µ 0 I 2 8π 2 1 r 2. The volume element is dv = l 2πr dr, so we have U = u m dv = µ 0 I 2 4π l b dr = µ 0 I 2 a r 4π l ln(b/a). d. Comparing to U = 1 2 LI 2, we see that L = µ 0 l 2π ln(b/a).

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