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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
xxxxxx0000
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]