Exam 3 NOV. 15.2018 - Physics 106 -R. Schad

YOUR NAME

1.

2.(7

3.

4.

5.

This is to identify the exam version you have — IMPORTANTMark the AThis is to identify the exam version you have — IMPORTANTMark the B

A cable with current flowing up, An electron, right of the cable, flies away from the cable.

The electron will feel a force in which direction?

a) Zero force

b)

d)

e)

A long, straight wire carries a current flowing right.A rectangular conducting loop lies in the same plane as the wire.

In order to induce a current flowingclockwise in the loop, you have to

/&DUCGP

a. Move the loop towards the wire.b. Move the loop away from the wire.

toll?c. Move the loop parallel to the wire to the rightd. Move the loop parallel to the wire to the lefte. impossible to get that done.

RIL = 5 H; = 21 V

The very first moment after the

switch was closed,

the voltage across R2 is

a) Zero V

c) 14 V

d) 18 v

e) 21 v

6.

L = 5 H; AVBattery -21 - V

After the switch was closed and wewaited a long time,the voltage across R2 is

a) Zero V

c) 14 vd) 18 ve) 21 v

7. Two long straight wires cross at the origin,each carrying a current of 1 A in the directionshown.

At which of the following points must themagnetic field due to the 2 wiresbe NOT zero?

A, B, C, and DB. A and Cc. A and DD. B and CE. B and D

8. A square loop of wire has sides of length 5 cm, and is completely placed in amagnetic field with magnitude given by B = 10t2, where B is in Tesla, t is inseconds. The normal of the loop area and B are parallel and pointing out of thepage. At t= 4 s the magnitude of induced voltage is:

1) 100 mV

2) 400 mV

3) 200 mV

4) Zero

5) Something else e 93

2S o /ö

9. Three long wires parallel to the x axis carry currents.If the upper wire (at y = 2 m) has 4 A to the right,the middle (at y = -1 m) 3 A to the left,the lowest (at y —-3 m) 1 A to the right.Whåt is the net force on a 1 m long segment of the middle cable?

-7

3

31

2a)Zerob) 11 x 10-7 N;c) 5 x 10-7

d) 11 x 10-7 N; Ve) 5 x 10-7

10. A wire of length 1 m carries a current of 10 A in the +Y direction through amagnetic field with components BX = +2T; By = +4T; BZ =0.The magnetic force on the wire is

a) Fx=20 N; Fy=40 N; Fz=ONb) Fx=o N; N; Fz=-20 N

d) Fx=o N; 6=20 N; Fz=-20 Ne) None of these

11. A charged particle flies Vand enters the regions with magnetic field as indicated below.In each case the particle is supposed to feel a force pointing to the right (9).The charges must be positive (+) or negative (-) ?

a) Qa +, Qb +, Q +

d) Qa Qb +, Q impossible

e) Qa +, Qb -, Q impossible

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xxxxxx0000xxxxxx

09

12. A long straight wire, perpendicular to the page passesthrough a uniform magnetic field [with 3 T strength]which points to the right.The net magnetic field values at points 1 and 2 areBI-net and B2-net •

Points I and 2 are the same distance from the wire.If IBI-netl > IB2-netl , then

l) the curent must flow @2) the curent must flow @3) the curent must be zero4) IBI-netl > IB2-netl cannot be true for any current in

the cable.

13. The figure shows three pairs of parallel plateswith the same separation of 1 m, and theelectric potential of each plate.Placing a positive +1 C charge in the center ofthe electric field for each of the 3 cases would

UniformB-ficld

2

1

Wire

9

result in forces on the charges like: —200 V —400 V -20 v +200[comparing magnitude and showing direction] (1)

a) I Fil > IF21 > IF31;b) I Fll > I h I > IF31;

c) IF21 > I Fil = I F31; d) > IFII=e) Something else

F3

F3

9

(3)

14. The point P lies along the perpendicular bisector of the line connecting two longstraight wires S and T perpendicular to the page. A set of directions A through H isshown next to the diagram. When the two equal currents in the wires aye directed oneup out of the page the other down into the page, the direction of the magnetic field at Pis closest to the directiop of

@T cD

G

10

15. In a uniform magnetic field which point to the right you place either a positive or anegative charge at rest.Which statement is correct?

1. The positive charge will experience a force to the left, the negative charge to the right.2. Both positive and negative charge will experience a force to the left.3. The positive charge will experience a force to the right, the negative charge to the left.4. Both positive and negative charge will experience a force to the right.5. Both positive and negative charge will experience no force.

6716. Three charges are placed at the corners of an equilateral triangle as shown. The charges

are equal in magnitude, but differ in sign as shown. What direction is the force on the

l)charge at the top of the triangle?

77

17. The figure shows the equipotential lines of an

electric "landscape". The numbers show thelevel of the electric potential in Volt.

A positive +1 C charge placed either at point Aor point B would feel a force [direction andrelative magnitude] like:

1) FAO > FBV

2) FAV > FB'T3) FAO < FBV

4) FAV < FB'fr

5)

93

o

Kinematics

Newton's Law

Conservation of Energy

Energy

Work

Power

(electrical)

Coulomb force

Electric field

Electric flux

Gauss Law

Potential energy

Potential

Capacitance

Charging/discharging

of Capacitor

x = xo + vo•t +

v2 ¯ v02 + — xo)

v

F

— 9.80 m/s2

KEI + UI + KE2 +

Kinetik (linear): KEIin = 1/2 mv2

Potential (gravity): Ug m g y

W-F.d=F d coso

P W/t=E/t

P = 1

F ke qiq2 / r.2

E-F/q=ke

12 R = (AV)2 / R

along the connecting line

dq2

volume

E keq/r2 for a point charge / pointing radially

E = —gradient(V) =dx

surface

E = (linside/€o = 4Ttke • qinside

closedsurfice

AU -UB-UA= —q] Eds = q AV

AV AU/q=

V = keq/r for a point charge

C = Q/AV [ = A/d parallel plate C]

ceq = CI + C2 + . [parallel combination]1/Ceq = I/CI + + l/C3 + ... [series combination]

U = Q2/2C = h Q h c (AV)2 [energy stored in C]

tfRC)

I(t) = (AV/R) •

q(t) = Q • e t"RC

¯ - AV/R • e-t/RC

Ohm's Law

Resistivity / Resistance

Magnetic force

Force between parallel currentcarrying cables

Cyclotron motion[circular motion of a charge in ama netic field

Magnetic Field by a current

Ampere's Law

Magnetic Flux

Faraday's Law of Induction

Lenz' Law

Inductor

LR circuit

electron mass

proton mass

elementary charge

Coulomb constant

Permittivity of free space

Permeability of free space

R = AV/I - = resistivi

p = me / (n q2 t) [T scattering time]

p = po [l + — To)] [temperature dependence]

1/Req= I/RI + l/R2+ 1/R3+ ... [parallel]

— RI + R2 + R3 + . series

FB=1.LxB [force on straight conductor]

force on conductor se ment

attractive force for same current directions.

repulsive force for opposite current directions.

* length2m

r = (mv) / (qB)

o = (qB) / m

[radius]

[frequency]

dB = go/4Tt • (I ds x r) / r2 [Biot-Savart Law]

B = (POI) / 2Tta) [long straight wire]

closedloop

J BdAsurface

emf= (AV • (dØB/ dt)

the polarity of the induced emf is such that it wants to act 'against' thecause.

emf= -L • dl/dt- e-t/i)

1 = AWR •

me = 9 x 10 31 kg

mp = 2 x 10 27 kg

e = 1.6 x 10-19 C

ke=9x 109 Nm 2/C 2

€0 = 9 x 10-12 C 2/Nm2

= 47t x 10-7 H/m

'switch on''switch off

[electron: -e ; proton: +e]

[ke = 1/4Ttto]