Electrostatics and Magnetism: Electric Fields, Charges, and Magnetic Fields, Lecture notes of Law

Download Electrostatics and Magnetism: Electric Fields, Charges, and Magnetic Fields and more Lecture notes Law in PDF only on Docsity! Practice problems for the final exam (with solutions) There will be at least a couple of problems on the final which will be identical to, or at least very similar to, problems listed here. Not all the questions here are multiple choice. Useful coefficients: g = 9.8 m s−2; Unit of elementary charge e = 1.6 × 10−19 C Mass of proton, mp = Mass of neutron, mn = 1.67 × 10−27 kg Mass of electron, me = 9.11 × 10−31 kg Coulomb’s constant, ke = 8.99 × 109 N m2 C−2 Permittivity of free space ǫ0 = 1 4πke = 8.85 × 10−12 C2 N−1 m−2 1 eV = 1.6 ×10−19 J 1 T = 104 G Permeability of free space µ0 = 4π × 10−7 T m A−1. Prefixes: milli (m) = 10−3; micro (µ) = 10−6; nano (n) = 10−9; pico (p) = 10−12; fempto (f) = 10−15 19-1: Excess charge on any conductor a. is always evenly distributed. b. is in the center. c. is on the surface. d. is fixed in space. e. cannot be detected. ANSWER:C 19-2: Electric field lines start a. at positive charges or at infinity. b. at negative charges or infinity. c. at infinity and end at positive charges. d. midway between positive charges and negative charges. e. midway between positive charges and infinity. ANSWER: A 19-3: The electric flux through a surface of fixed area is maximum when the surface is a. parallel to the electric field. b. antiparallel to the electric field. c. perpendicular to the electric field. d. at an angle of 45◦ to the electric field. e. closed, but does not contain the charge. ANSWER: C: Perpendicular. Note that this holds for any arbitrary E-field or surface– as long as the local electric field is perpendicular to the surface at a given location, then ΦE is maximized at that location (maximum number of field lines penetrating the surface). ΦE = E A cos(θ), and this is maximized when cos(θ)=1 and θ=0◦. If ~E can be made to be perpendicular to the surface at all points on the surface, then the sum total ΦE will be maximized. 1 19-4: When a conductor is in electrostatic equilibrium, the electric field inside the conductor is . (Choices: 0, σ/ǫ0, 2σ/ǫ0, σ/(2ǫ0) ). ANSWER: E-field inside a conductor at equilibrium is zero. 19-5: When a conductor is in electrostatic equilibrium, the electric field a point just outside the conductor is . (Choices: 0, σ/ǫ0, 2σ/ǫ0, σ/(2ǫ0) ). ANSWER: E-field just outside a conductor is σ/ǫ0. (Compare to the case of an infinite plane of uniform charge, where both above & below the plane, the magnitude of ~E is σ/(2ǫ0)) 19-6: The xy plane is ”painted” with a uniform surface charge density of 40 nC/m2. Consider a spherical surface with a 0.04m radius that has a point in the xy plane as its center. What is the total electric flux ΦE in Nm2/C? ANSWER: Use Gauss’ Law: ΦE = Qencl/ǫ0. The amount of charge enclosed by the sphere will be that contained on a circle of radius 0.04m, since that’s the area inside the Gaussian spherical surface. Qencl = π(0.04m)2 × 40 nC/m2 = 2.0×10−10 C. ΦE = Qencl/ǫ0 = (2.0×10−10 C) / (8.85 × 10−12 C2 N−1 m−2) = 22.7 Nm2/C 20-1. The electrical potential energy of a pair of like charges is a. positive. b. negative. c. zero. d. proportional to the square of the distance. e. inversely proportional to the square of the distance. ANSWER: U = keq1q2 r . If both q1 & q2 are positive, U is positive. If both q1 & q2 are negative, U is again positive. For a pair of charges, U will be negative only when the charges have opposite signs. So the answer here is A. Choice E is rejected because U and V are both proportional to 1/r, and the magnitudes of ~F and ~E are inversely proportional to the square of the distance. 20-2. Equipotential surfaces are a. perpendicular to electric field lines. b. parallel to electric field lines. ANSWER: perpendicular: see section 20.2 2 Notice that ~F13 = – ~F31, ~F12 = – ~F21, and ~F32 = – ~F23; in each case, vectors are equal in magnitude but opposite in direction. To determine the amount of work needed to assemble this 3-particle system by moving the points in from infinity, we must calculate the system’s total potential energy, considering all three pairs of interactions: Utot = U12 + U23 + U13 Utot = keq1q2 r12 + keq2q3 r23 + keq1q3 r13 Utot = (ke)(+3µC)(−2µC) 0.1m + (ke)(−2µC)(+8µC) 0.15m + (ke)(+3µC)(+8µC) 0.25m Utot = (–0.54J) + (–0.96J) + (+0.86J) = –0.64 J. Note that because the potential energy is negative, this is a relatively stable system. To assemble it, we (as an external agent) don’t even have to put in any work; the electric fields can do all the work (+0.64J of work, to be exact) for us in this case. 21.1. A 9.0-volt battery moves 100 milliCoulombs of charge through a circuit connecting the battery’s terminals. How much energy was delivered to the circuit? a. 90.0 J b. 9.0 J c. 0.9 J d. 3.0 J e. 0.3 J The energy can calculated via ∆PE = q∆V = (0.1 C)(9.0 V) = 0.9 J. 21.2: Find Req and the magnitude of the currents through each resistor. ANSWER: Solution: The three resistors in the right branches are all connected in parallel. 1 Req = 1 5Ω + 1 10Ω + 1 5Ω = 5 10Ω = 5 10Ω Req1 = 2Ω We then take this resistance and combine it in series with the 5Ω resistor on the bottom to obtain Req for the whole circuit: Req = 2Ω + 5Ω = 7Ω Let’s call Itot the current through the battery, and I1, I2, & I3 the currents through each of the branches on the 5 right side; in accordance with the Junction Rule, Itot = I1 + I2 + I3. Now we can use ǫ = ItotReq to solve for Itot: Itot = 21V/7Ω = 3A. To solve for I1, I2, & I3, we note that the voltage drop across each of the resistors in the right branch is identical – but it will not be equal to 21V. In accordance with the loop rule: It will be equal to 21V minus the ∆V associated with traversing the bottom 5Ω resistor. That ∆V is (5Ω)(Itot) = (5Ω)(3A) = 15V. The voltage drop across each branch is thus 6V. We now solve for the current in each branch: The current flowing through the 10Ω resistor is ∆V/R = 6V/10Ω = 0.6 A. The current flowing through each 5Ω resistor on the right side is ∆V/R = 6V/5Ω = 1.2 A. Notice that the Junction Rule, Itot = I1 + I2 + I3, is confirmed: 3A = 1.2A + 1.2A + 0.6 A! 22-1. The magnetism of ferromagnetic materials, such as iron, has its origin in the intrinsic magnetic moment of electrons that is associated with their a. orbital angular momentum. b. spin angular momentum. c. electric charge. d. being magnetically soft. e. being magnetically hard. ANSWER: B (see section 22.11) 22-2. When two parallel current-carrying wires have currents in the same direction, the wires a. attract one another. b. repel one another. c. exert no forces on one another. d. exert equal and opposite repulsive forces on one another perpendicular to the plane of the wires. e. exert equal and opposite attractive forces on one another perpendicular to the plane of the wires. ANSWER: A (see section 22.8) 22-3 The head end of a nail is attracted to the north pole of a magnet and also to its south pole. a. The nail is magnetized. b. The nail is initially not magnetized. c. The nail is aluminum. d. The nail is copper. e. The magnet is aluminum. ANSWER: B – the nail must be composed of a soft magnetic material, since it responds to whatever B-field it is placed in (near either pole of a magnet), then loses its magnetism when taken away from the pole of the magnet. 6 22-4. The head of a nail is attracted to the north pole of a magnet and repelled by the magnet’s south pole. a. The nail is magnetized. b. The nail is initially not magnetized. c. The nail is aluminum. d. The nail is copper. e. The magnet is aluminum. ANSWER: A – the nail must be a permanent magnet, comprised of a hard magnetic material, and having its own permanent magnetic field. 22-5. If an electron moves from right to left across this page and suddenly is subjected to a magnetic field directed into the page, what is the direction of the magnetic force on the electron? a. Into the page b. Out of the page c. From the bottom to the top of the page d. From the top to the bottom of the page e. From the left to the right of the page ANSWER: C. Right hand rule, for a positive charge: Thumb = ~v = left. Fingers = ~B = into page. Out of palm = ~F = down. Negative charge: reverse direction. 22-6: An electron is located on the Earth’s magnetic equator at an altitude of several km. In which direction can the velocity vector be pointing such that the deflection due to magnetic force is zero? Choices are: magnetic north, upward, downard, magnetic south, magnetic east, magnetic west. Note that 2 answers are acceptable. ANSWER: If ~v and ~B are parallel to each other, then |~F | = 0. At the magnetic equator, the local magnetic field lines are oriented magnetic-north-south. Answers = magnetic N & S. 22-7: Suppose you have in your possession a gigantic single copper loop that has a radius r of 1.1 km. You want to run a current through the loop to generate a magnetic field whose strength at the center of the loop is ten times the strength of the Earth’s magnetic field at the surface (roughly 0.3 to 0.5 G, depending on location. Let’s choose 0.3 G). The total resistance of the loop is 0.12Ω. A: What EMF do you need to supply? B: If the loop is horizontal, in which direction should the current flow if you want the B-field generated by the current loop to be directed downward? The loop is made from pure copper, which, as per Table 21.1, has a resistivity ρ of 1.7×10−8Ωm at an operating temperature T0=20◦C. C: The loop has a square cross section. What is the cross-sectional area of the loop and the size of the square? D: How much power is dissipated in the resistor? E: Suppose a friend of yours (who has not taken Physics 1B yet) thinks that making the loop hotter will “somehow” make the magnetic field strength larger. He/she heats up the loop uniformly (I dunno, a giant blowtorch?), so that its temperature is 80◦C. Assume the EMF supplied does not change. What effect will this really have on the B-field at the center 7 23-3: A 100-turn coil with cross-section 0.1 m2 is immersed in a magnetic field of 0.01 T. Initially, the plane of the loops is perpendicular to the magnetic field lines. Then, in a time ∆t, the coil is turned 90◦ such that the magnetic flux is zero. You would like to obtain an induced EMF in the coil whose magnitude is 2 V. Calculate the ∆t necessary to accomplish this. ANSWER: From Faraday’s Law: ǫind = −N ∆ΦB ∆t ∆ΦB = B∆A. ∆A = A − 0 Rearrange Faraday’s Law to solve for ∆t (and let’s drop the negative sign since time cannot be negative, and we were just given the magnitude, not direction, of the induced emf) : ∆t = (N∆ΦB)/ǫind = ((100)(0.01T)(0.1m2))/(2V) = 0.05 s. 23-4: A bar magnet approaches, and passes through, a con- ducting loop from the left with its north pole directed to- wards the right. It moves with a constant velocity; the loop, which has radius r, is held stationary. The current in the loop has its greatest magnitude AND a clockwise direction a. as the north pole is approaching the loop, a distance 3r away b. just as the north pole of the magnet is entering the loop c. When the magnet is exactly halfway through the loop d. as the south pole is leaving the loop e. after the south pole has left the loop and is a distance 3r away ANSWER: The CHANGE in magnetic flux is maximum when one of the poles of the magnet is passing through the loop. When the north pole is first entering the loop, the magnet’s B-field lines are pointing from left to right and the magnetic flux through the loop is increasing. Consistent with Lenz’s law, the induced B-field will try to counteract this increase, and will be oriented right-to- left. The induced current in the loop needed to generate right-to-left induced B-fields will go counterclockwise. Later, when the south pole is passing through the loop, the magnet’s B-field lines are still pointing from left to right, though now the magnetic flux through the loop is decreasing. Consistent with Lenz’s law, the induced B-field will try to counteract this decrease, and will be oriented left-to-right to attempt to bolster the decreasing original field. The induced current in the loop needed to generate left-to-right induced B-fields will go clockwise. Final answer = D. 23-5. An ideal transformer is one that: a. has a turn ratio, N2/N1, equal to 1. b. works with direct current. c. experiences no power loss. d. has an output frequency of 60 Hz. ANSWER = No power loss. 10 23-6. An AC power source produces a peak voltage of 7200 V and a peak current of 100 A in a circuit. This circuit is attached to an ideal transformer whose job is to step-up the voltage to 200,000 V (peak) for energy transmission in a long-distance power line. What is the peak current in the long-distance line? a. 1.0 A b. 2.8 A c. 24 A d. 2800 A e. 3.6 A Solution: Power input at primary = power output at secondary: I1∆V1 = I2∆V2 I2 = I1∆V1/∆V2 = (100 A)(7200 V)/(2×105 V) = 3.6 A 11