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Phys 0175 Practice Midterm Exam III Solutions Apr 1, 2009
Note: THIS IS A REPRESENTATION OF THE ACTUAL TEST. It is a sample and does not include questions on every topic covered since the start of the semester.
Also be sure to review homework assignments on WebAssign White board problems worked in the class Exercises, Examples, and Review Questions (at the end of each chapter) in your textbook On the actual test, do not use other paper. If you need more space, write on the blank page included at the beginning of the test, and indicate that you did this.
- Read all problems carefully before attempting to solve them.
- Your work must be legible, and the organization must be clear.
- You must show all your 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 complete but brief. Do not write a lot of prose.
- Include diagrams!
- Show what goes into a calculation, not just the final number: a · b c · d =
(8 × 10 −^3 )(5 × 106 )
(2 × 10 −^5 ))(4 × 104 ) = 5^ ×^10
4
- 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.
NAME:
Phys 0175 Practice Midterm Exam III Apr 1, 2009
INSTRUCTIONS: Please print your name above in the space provided.
The exam consists of N questions worth differing number of points. WRITE NEATLY. Clearly mark your answers with a bounding box at the bottom of your work area. It is important to show your work to get credit. You may use calculators.
Question Number Possible Points Score 1? 2? ... ... N? Total 100
- The wires in the circuit shown in the diagram are made of the same material, but one segment of wire is thinner than the other wires.
B
C
x
x x
x
x x x
E
E
E
E E (^) E (^) v E
v
(a) At each location inside the wire, marked by an “x”, draw and label an arrow representing the electric field at that location. Relative magnitudes of the arrows should be correct. Label the arrows “E”. (b) At locations B and C, draw and label arrows representing the drift speed of an electron at that location. Label the arrows “v”. (c) Draw an approximate distribution of charge on the surface of the wires. Make sure the distribution you show is consistent with your answer to (a).
- A particular resistive metal has 3× 1027 mobile charges per cubic meter, with a mobility of 4. 2 × 10 −^5 (m/s)/(V/m). When a metal bar made of this material (bar a), 7 cm long, diameter unknown, is put into a circuit (circuit 1) with one battery whose emf is 1.5 V, and thick connecting wires, 1. 2 × 1017 electrons per second enter bar a. When a different metal bar (bar b) made of the same material, 5.5 cm long, with a different diameter, is put into a different circuit (circuit 2), with one battery whose emf is 1.5 V, and thick connecting wires, 2. 7 × 1017 electrons per second enter bar b. - +
Circuit 1
bar a
7 cm
Circuit 2
bar b
5.5 cm
Circuit 3, shown in the diagram below, contains two batteries, each with emf 1.5 V, thick connecting wires, and three resistors: one resistor identical to bar a (same material and dimensions as bar a) and two resistors identical to bar b (same material and dimensions as bar b).
bar b bar b
X
X X
(a) Calculate the magnitude of the electric field inside bar a in circuit 3. Clearly show all steps in your work. For the first and second circuits, we have (EMF = E) E = EL =⇒ E = (^) LE, i = nAuE =⇒ A = (^) nuEi = (^) nuiLE. Therefore, Aa = 1.^2 ×^10
3 × 1027 · 4. 2 × 10 −^5 · 1. 5 m
(^2) = 4. 4 × 10 − (^8) m (^2) ,
Ab = 2.^7 ×^10
3 × 1027 · 4. 2 × 10 −^5 · 1. 5 m
(^2) = 7. 9 × 10 − (^8) m (^2). We also have 2E − EaLa − 2 EbLb = 0 and nAauEa = ia = ib = nAbuEb for circuit
- Therefore, Eb = A Aab Ea =⇒ Ea = (^) La + 2^2 LEaAa/Ab = (^0) .07 + 2 ·^2 0 ·.^0551.^5 · 4. 4 / 7. 9 V/m = 22.9 V/m. (b) The locations marked “x” on the diagram are inside the metal bars. At each “x” in the diagram of circuit 3, draw an arrow indicating the direction and relative magnitude of the electric field at that location. (Eb = EaAa/Ab = 12.8 V/m .)
- A capacitor, which is initially uncharged, is connected to a battery by a thing Nichrome wire, as shown in the diagram. A compass lying under the wire originally points North, but when the circuit is connected the compass needle deflects away from North. After 20 seconds, the compass needle again points North.
(a) On the axis below, draw a graph showing the magnitude of the net electric field at location B (marked by an “x”) inside the Nichrome wire, as a function of time. The circuit (batteries, wire, uncharged capacitor) is connected at time t = 0.
-^ +
x^ B
North
0 10 s 20 s time
| E^
| at location
B
net
(b) The expanded diagram at right shows location B at t = 30 s. On the diagram draw and label three arrows representing
- The electric field E~cap at that location due to the charges on the capacitor plates
- The electric field E~surf at that location due to the charges on the battery and the surface of the wires
- The net electric field E~net at that location If any quantity is zero, state this explicitly. The directions and relative magni- tudes of the arrows should be correct.
B
Esurf (^) x Ecap E (^) net = 0
- A capacitor consists of two circular plates, each 1.5 m in radius, separated by a 2 mm gap. The capacitor is charged and is connected to a resistive wire 5 m long, as shown in the diagram, and current runs through the wire. At a particular instant, the magnitude of the electric field at the location inside the wire marked by “x” is 3.5 V/m. At this instant, how much charge is on the positive plate of the capacitor? Show your work.
L = 5 m x
2 mm
R = 1.5 m
The electric field is constant along the wire. We can use the loop equation:
∆Vcap − EL = 0 =⇒ ∆Vcap = EL = 3.5 V/m × 5 m = 17.5 V.
Since the electric field is almost constant inside the capacitor, we can calculate the electric field there:
∆Vcap = Eind =⇒ Ein = ∆V dcap = (^2) ×^17 10 .5 V− (^3) m = 8750 V/m.
The electric field inside a capacitor is given by
Ein = Q/A 0.
Solving for Q,
Q = 0 AEin = 8. 85 × 10 −^12 C^2 /(N · m^2 ) × π(1.5 m)^2 × 8750 N/C = 5. 47 × 10 −^7 C.
- A long straight wire lies on a table, oriented along the x-axis. A conventional current of 9 A runs through the wire in the +x direction. At a particular instant, an electron traveling at a speed of 3 × 107 m/s in the +x direction passes 2 mm above the center of the wire. What is the force on the electron at this instant? Your answer must be a vector. Show your work. If you made any assumptions or approximations, state them.
y -
x
v
2 mm
wire
The force on the electron is F^ ~ = q~v × B,~ so we need to figure out B~ at the location of the electron. Using the right-hand rule, the direction of the magnetic field is in the +z direction. The magnitude is
| B~| = 4 μπ^02 rI = 10−^7 T^ A·^ m 2 ×^2 × 10 9 A− (^3) m = 9 × 10 −^4 T,
where I have used the approximation that we have an infinitely long wire. The force is then
F^ ~ = − 1. 6 × 10 −^19 C · 3 × 107 m/s · 9 × 10 −^4 T(ˆx × ˆz) = − 4. 32 × 10 −^15 N(ˆx × zˆ).
The cross product is xˆ × ˆz = −ˆy, so F^ ~ = 4. 32 × 10 −^15 N ˆy = 〈 0 , 4. 32 × 10 −^15 , 0 〉 N.
- A rectangular copper bar with length d = 0.55 m, height = depth = w = 0.02 m, is pulled upward by a rope (which is not shown), at a constant speed v = 0.3 m/s, through a region in which there is a uniform magnetic field of 0.43 tesla, directed into the page. The copper bar slides along metal rails, with negligible friction. (You may not need to use all the information given in this problem.) In the following questions, if any
quantity is zero, state this explicitly. R
d
w
B v
conventional current
(a) On the diagram, draw pluses and minuses indicated the charge distribution in and/or on the bar. (b) On the diagram, draw an arrow indicating the direction of conventional current through the resistor, and label it “conventional current”. (c) What is the absolute value of the potential difference between the ends of the bar? Start from fundamental principles (do not start with a formula that is not on the formula sheet). Show all steps in your work. When the charges are all built up, the magnetic force will compensate the electric force. We therefore have
F^ ~ = q( E~ + ~v × B~) = 0 =⇒ | E~| = |~v × B~|.
Since the magnetic field is perpendicular to the velocity, we have
E = vB = 0.3 m/s × 0 .43 T = 0.129 V/m.
The potential difference is then
∆V = Ed = 0.129 V/m × 0 .55 m = 0.071 V.
(e) What is the drift speed v of the mobile charges? Start only from principles, relations, and equations on the formula sheet. Explain your reasoning using words, equations, and/or diagrams. The magnetic force on the charges is canceled by the electric field E~⊥. We therefore have E^ ~⊥ + ~v × B~ = 0. The velocity is perpendicular to B~, and the cross product is in the same direction as the perpendicular electric field. We therefore have
¯v = |^
E~⊥|
| B~
= ∆V /L
| B~|
=^0.^045 ×^10
− (^3) V/ 0 .04 m 0 .85 T = 1. 32 × 10 −^3 m/s.
(f) What is the mobility u of the mobile charges?
The mobility is related to the drift speed and the electric field by
u = E ¯v.
In this case, the electric field is E‖, which is
E‖ = ∆LV = (^02) ..25 m 7 V = 10.8 V/m.
The mobility is then
u =^1.^32 ×^10
− (^3) m/s 10 .8 V/m = 1.^22 ×^10
− (^4) (m/s)/(V/m).
- The electric field is measured all over a box- shaped surface, as shown in the diagram. What is the net charge inside the box? Show all steps in your work, and explain how you evaluated any quantities related to angles or dot products. If any quantity is zero, state this explicitly.
E 1 = 3000 V/m
E (^) 2 = 1500 V/m
E (^) 3 = 400 V/m
Gauss’ law is (^) ∮ E^ ~ · ˆndA = Qinside 0 Since the electric field is constant on the surfaces, the integral is easy. On the sides, the unit normal is perpendicular to the electric field, so the contribution to the flux from the sides is zero, Φsides = 0. On the bottom surface, the unit normal points down, while the electric field points up, so the contribution to the flux from the bottom is
Φbottom = −E 1 A = −3000 V/m × 0 .20 m × 0 .03 m = −18 V · m.
On the top surface, both the electric field and the unit normal are in the same direction, so the flux from the top is
Φtop = −E 3 A = −400 V/m × 0 .20 m × 0 .03 m = 2.4 V · m.
The total flux is then
Φtop + Φbottom + Φsides = 2.4 V · m − 18 V · m + 0 = − 15 .6 V · m.
The net charge inside is then
Qinside = 0 Φ = 8. 85 × 10 −^12 V C· m × (− 15 .6 V · m) = − 1. 38 × 10 −^10 C.
- Here is a pattern of magnetic field. Calculate the amount and direction of the current passing through the shaded region. Show all your work. 1 T 1 m 5 m
1 T 1 T 1.5 T 1.5 T
2 T 2 T 2 T
There is a net circulation in the counter-clockwise direction, which using the right- hand rule tells us that the current is flowing out of the page. Using Ampere’s Law, we have (^) ∮ B^ ~ · d~l = μ 0 Iinside. Calculating the integral on the left, we have 4 contributions, from each side. The two vertical pieces give zero, since d~l is vertical, while B~ is horizontal. The bottom piece gives (^) ∫ bottom B^ ~ · d~l = 2 T × 5 m = 10 T · m. The top piece gives (^) ∫
top
B^ ~ · d~l = −1 T × 5 m = −5 T · m, where the minus sign comes from the dot product (d~l now points to the left, B~ to the right). We therefore have (^) ∮ B^ ~ · d~l = 5 T · m. This is equal to μ 0 Iinside, so
Iinside = 5 T μ^0 ·^ m = (^4) π × 10 5 T− 7 · (^) Tm · m/A = 3. 98 × 106 A.
Things you must know
Relationship between electric field and electric force Conservation of charge Electric field of a point charge The Superposition Principle Magnetic field of a moving point charge
Other Fundamental Concepts
∆Uel = q∆V ∆V = −
∫ (^) f i
E^ ~ · d~l ≈ −Σ(Ex∆x + Ey∆y + Ez ∆z) Φel =
E^ ~ · n dAˆ Φmag =
B^ ~ · n dAˆ ∮ E^ ~ · n dAˆ =
∑ (^) q inside 0
B^ ~ · n dAˆ = 0
Ampere without Maxwell (no displacement current) ∮^ B~ · d~l = μ 0 ∑^ Iinside path
Specific Results
E^ ~ due to uniformly charged spherical shell: outside like point charge; inside zero
| E~dipole, axis| ≈ (^4) π^102 rqs 3 (on axis, r s) | E~dipole, ⊥| ≈ (^4) π^10 qsr 3 (on ⊥ axis, r s)
| E~rod| = (^4) π^10 r√r (^2) + (^ QL/2) 2 (r ⊥ from center) | E~rod| ≈ (^4) π^102 Q/Lr (if r L)
Electric dipole moment p = qs | E~ring| = (^4) π^10 (z (^2) +^ qz R (^2) ) 3 / 2 (z alongaxis)
| E~disk| = Q/A 2 0
[
1 − √z 2 z+ R 2
]
(z alongaxis) | E~disk| ≈ Q/A 2 0
[
1 − (^) Rz
]
≈ Q/A 2 0 (if z R)
| E~capacitor| ≈ Q/A (+Q and − Q disks) | E~f ringe| ≈ Q/A
( (^) s 2 R
(just outside capacitor)
∆ B~ = 4 μπ^0 I∆
~l × ~r r^2 (shortwire) | B~wire| = 4 μπ^0 r√r 2 LI+ (L/2) 2 ≈ μ 4 π^02 rI (r L)
| B~loop| = 4 μπ^02 IπR
2 (z^2 + R^2 )^3 /^2 ≈^
μ 0 4 π
2 IπR^2 z^3 (on axis, z^ ^ R)^ μ^ =^ IA^ =^ IπR
2
| B~dipole, axis| ≈ μ 4 π^02 rμ 3 (on axis, r s) | B~dipole, ⊥| ≈ 4 μπ^0 rμ 3 (on ⊥ axis, r s)
