2212_Test3 - PHYS 2212 Test 3 Nov 10th 2010 Name(print)...

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Unformatted text preview: PHYS 2212 Test 3 Nov 10th 2010 Name(print) Instructions • Read all problems carefully before attempting to solve them. • Your work must be legible, and the organization must be clear. • You must show all work, including correct vector notation. • Correct answers without adequate explanation will be counted wrong. • Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything you don’t want us to read! • Make explanations correct but brief. You do not need to write a lot of prose. • Include diagrams! • Show what goes into a calculation, not just the final number, e.g.: 5 × 104 a·b c·d = (8×10−3 )(5×106 ) (2×10−5 )(4×104 ) = • Give standard SI units with your results. Unless specifically asked to derive a result, you may start from the formulas given on the formula sheet, including equations corresponding to the fundamental concepts. If a formula you need is not given, you must derive it. If you cannot do some portion of a problem, invent a symbol for the quantity you can’t calculate (explain that you are doing this), and use it to do the rest of the problem. Honor Pledge “In accordance with the Georgia Tech Honor Code, I have neither given nor received unauthorized aid on this test.” Sign your name on the line above PHYS 2212 Do not write on this page! Problem Problem Problem Problem Problem Score 1 2 3 4 (25 (25 (25 (25 pts) pts) pts) pts) Grader FINAL EXAM CONFLICT RESOLUTION The final exam period has been set by the Registrar’s Office. The Final Exam for all sections of 2212 will be held during Period Seven, on Wedesday December 15, from 8:00 AM to 10:50 AM. If you think you have a conflict please read the full announcement posted on WebAssign and email Dr. Greco the requested information. The departmental deadline for notifying us about a potential conflict is Monday, November 15! Problem 1 (25 Points) A compass is located at the center of a square coil of wire consisting of 3 turns. Each side of the square is 6 cm long. Originally, the compass points North, in the −x direction. Then the wire is connected to a battery (which is not shown), and a conventional current of 0.45 A runs in the coil, in the direction indicated on the diagram. (a 5pts) What direction does the compass needle point while the current is running? Draw the direction clearly on the diagram below (which is a top view of the compass). N (b 8pts) Determine the magnitude of the compass needle’s angular deflection while the current is running. Clearly show each step in your work. (c 7pts) Determine the magnetic dipole moment of the square coil. Show all steps in your work. (d 5pts) Compute the magnitude of the magnetic field at location 0, 1.5, 0 m due only to the current in the coil. Show all steps in your work. Problem 2 (25 Points) A horizontal pipe contains a dilute solution of sulfuric acid (H2 SO4 ). There are 4 × 1025 sulfate ions (SO2− ) 4 per m3 (charge -2e) and 8 × 1025 hydrogen ions (H+ ) per m3 (charge +e). An electric field is applied to the right along the pipe, and the hydrogen ions acquire a drift speed of 9 × 10−5 m/s while the sulfate ions acquire a drift speed of 2 × 10−5 m/s. The pipe has a cross-sectional area 2.4 × 10−4 m2 . Determine the magnitude and direction of the conventional current. Problem 3 (25 Points) A capacitor consists of two circular plates, each 1.5 m in radius, separated by a gap of 2 mm. The capacitor is charged and is connected to a resistive wire 5 m long, as shown in the diagram. A current runs through the wire. At a particular instant, the magnitude of the electric field at the location inside the wire marked by the “x” is 3.5 V/m. At this instant, determine how much charge is on the positive plate of the capacitor. Show your work. 2 mm R = 1.5 m L =5 m × Problem 4 (25 Points) A U-shaped piece of wire hangs from supports and is connected to a battery, as shown. A steady current of 4.5 A runs though the U-shaped wire. The bottom piece of the wire is 25 mm long, and is oriented along the z axis. The vertical sides of the U are 4 cm long. y 4 cm A coil of wire is placed on the −x axis, with its axis oriented along the x axis. When the coil is connected to a power supply that is not shown in the diagram, the 25 mm section of the Ushaped wire is acted on by a force in the +y direction 0, 6 × 10−3 , 0 N. 25 m m x z (a 10pts) If you stand on the +x axis and look toward the coil, which way is conventional current running in the coil? In addition to describing the direction in words, show it on the diagram. To receive full credit, explain briefly using diagrams that include vectors. (c 15pts) Determine the magnetic field at the location of the 25 mm section of the wire, due to the coil. Your answer should be a vector and be sure to show all steps in your work. This page is for extra work, if needed. Things you must know Relationship between electric field and electric force Electric field of a point charge Relationship between magnetic field and magnetic force Magnetic field of a moving point charge Conservation of charge The Superposition Principle Other Fundamental Concepts dp dp = Fnet and ≈ ma if v << c dt dt f ∆V = − i E • dl ≈ − (Ex ∆x + Ey ∆y + Ez ∆z ) Φmag = B • ndA ˆ dv dt ∆Uel = q ∆V Φel = E • ndA ˆ qinside E • ndA = ˆ a= 0 |emf| = EN C • dl = B • ndA = 0 ˆ dΦmag dt Iinside path + B • dl = µ0 B • dl = µ0 0 d dt Iinside path E • ndA ˆ Specific Results 1 2qs (on axis, r s) 4π 0 r 3 Q 1 (r ⊥ from center) Erod = 4π 0 r r 2 + (L/2)2 1 2Q/L Erod ≈ (if r L) 4π 0 r z Q/A (z along axis) 1− 2 Edisk = 20 (z + R2 )1/2 Q/A (+Q and −Q disks) Ecapacitor ≈ Edipole,axis ≈ 0 µ0 I ∆ × r ˆ ∆B = (short wire) 2 4π r µ0 µ0 2I LI Bwire = ≈ (r 2 + (L/2)2 4π r r 4π r 1 −qa⊥ 4π 0 c2 r i = nAv ¯ Erad = σ = |q | nu Edielectric = Eapplied K 1 qs (on ⊥ axis, r 4π 0 r 3 s) electric dipole moment p = qs, p = α Eapplied qz 1 (z along axis) 2 + R2 )3/2 4π 0 (z Q/A Q/A z Edisk ≈ ≈ (if z R) 1− 20 R 20 Q/A s just outside capacitor Ef ringe ≈ 2R 0 Ering = ∆F = I ∆l × B L) µ0 2IπR2 2IπR2 µ0 ≈ (on axis, z 4π (z 2 + R2 )3/2 4π z 3 µ0 2µ Bdipole,axis ≈ (on axis, r s) 4π r 3 Bloop = Edipole,⊥ ≈ Bwire = Bearth tan θ R) µ = IA = IπR2 Bdipole,⊥ ≈ µ0 µ (on ⊥ axis, r 4π r 3 ˆ ˆ v = Erad × Brad ˆ Brad = I = |q | nAv ¯ I J= = σE A 1 q 1 ∆V = − 4π 0 rf ri v = uE ¯ L R= σA due to a point charge s) Erad c I= | ∆V | for an ohmic resistor (R independent of ∆V ); R Q = C |∆V | power = I ∆V Power = I ∆V K ≈ 1 mv 2 if v 2 c I= circular motion: dp |v | mv 2 = |p| ≈ dt ⊥ R R |∆V | (ohmic resistor) R Math Help a × b = ax , ay , az × bx , by , bz = (ay bz − az by )ˆ − (ax bz − az bx )ˆ + (ax by − ay bx )ˆ x y z dx = ln (a + x) + c x+a dx 1 =− +c 2 (x + a) a+x a dx = ax + c Constant Speed of light Gravitational constant Approx. grav field near Earth’s surface Electron mass Proton mass Neutron mass Electric constant Epsilon-zero Magnetic constant Mu-zero Proton charge Electron volt Avogadro’s number Atomic radius Proton radius E to ionize air BEarth (horizontal component) ax dx = Symbol c G g me mp mn 1 4π 0 0 µ0 4π µ0 e 1 eV NA Ra Rp Eionize BEarth a2 x +c 2 dx 1 =− +c 3 (a + x) 2(a + x)2 ax2 dx = a3 x +c 3 Approximate Value 3 × 108 m/s 6.7 × 10−11 N · m2 /kg2 9.8 N/kg 9 × 10−31 kg 1.7 × 10−27 kg 1.7 × 10−27 kg 9 × 109 N · m2 /C2 8.85 × 10−12 (N · m2 /C2 )−1 1 × 10−7 T · m/A 4π × 10−7 T · m/A 1.6 × 10−19 C 1.6 × 10−19 J 6.02 × 1023 molecules/mole ≈ 1 × 10−10 m ≈ 1 × 10−15 m ≈ 3 × 106 V/m ≈ 2 × 10−5 T ...
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