Download PHYS 221 Exam 2 10 July 2015 Physics 221 – Exam 2 Lorentz

Physics 221 – Exam 2
𝑒 = 1.6 × 10−19 𝐶
𝑘 = 9.0 × 109
𝑁⋅𝑚2
𝐶
⃑)
Lorentz Force Law: 𝑞(𝐸⃑ + 𝑣 × 𝐵
∆𝑉 =
∆𝑃𝐸𝑒𝑙𝑒𝑐
𝑞
Q = 𝐶Δ𝑉
𝐼=
= −𝐸⃑ ⋅ ∆𝑟
𝑄2
𝐶2
𝜀𝑜 = 8.9 × 10−12 𝑁⋅𝑚2
⃑ ⋅𝐴
Φ𝐵 = 𝐵
point charge: 𝑉 =
1
1
𝑈𝐸 = 2𝐶 = 2 𝑄∆𝑉 = 2 𝐶(∆𝑉)2
Δ𝑄
Δ𝑡
Δ𝑉𝑅 = 𝐼𝑅
Δ𝐼
Δ𝑉𝐿 = 𝐿 Δ𝑡
For AC Circuits:
𝑃 = 𝐼𝑉
Φ = 𝐿𝐼
Δ𝑉𝐿 = 𝐼𝑋𝐿
|𝑍| = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
Δ𝑉𝐶 = 𝐼𝑋𝐶
𝜇 𝐼
𝑜
Long, Straight Wire: 𝐵 = 2𝜋𝑟
1 Volt = 1 J/C
𝐿
𝐶𝑠𝑒𝑟 = (∑ 𝐶 )
𝑖
𝑅𝑠𝑒𝑟 = ∑ 𝑅𝑖
1 −1
𝐿𝑝𝑎𝑟 = (∑ 𝐿 )
𝑖
𝑋𝐶 =
1
2𝜋𝑓𝐶
1 Volt/m = 1 N/C
1 −1
𝐶𝑝𝑎𝑟 = ∑ 𝐶𝑖
𝑅 = 𝜌𝐴
𝐿𝑠𝑒𝑟 = ∑ 𝐿𝑖
𝑘𝑞
𝑟
𝑚⋅𝑘𝑔
𝜇𝑜 = 4𝜋 × 10−7 𝑠2 ⋅𝐴2
1 −1
𝑅𝑝𝑎𝑟 = (∑ 𝑅 )
𝑖
𝜇𝑁 2 𝐴
𝑙
Solenoid: 𝐿 =
𝑋𝐿 = 2𝜋𝑓𝐿
|𝑉| = √(Δ𝑉𝑅 )2 + (Δ𝑉𝐿 − Δ𝑉𝐶 )2
𝐶 = 𝜅𝐶𝑣𝑎𝑐
𝐼𝑅𝑀𝑆 = 𝐼𝑃𝑒𝑎𝑘
𝑋𝐿 −𝑋𝐶
)
𝑅
𝛿 = ArcTan (
PHYS 221
Exam 2
THIS PAGE LEFT BLANK
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10 July 2015
PHYS 221
Exam 2
10 July 2015
1. An RLC series circuit consists of a 1 kΩ resistor, a 1 μF capacitor and a 0.2 H inductor connected
to a 150 V A/C source. What is the current delivered by the source at resonance?
a. 20.3 mA
b. 150 mA
At resonance, the impedance is equal to the resistance which
means the current is just
c. 10.3 mA
150 𝑉
1𝑘 Ω
= 150 𝑚𝐴
d. 15.0 mA
e. 54.3 mA
2. Consider the circuit shown below with V = 9.0 V, R1 = 3000 Ω, R2 = 7000 Ω and L = 7.7 mH. What
is the energy stored in the inductor?
a. 3.5 x 10-8 J
b. 6.9 x 10-8 J
c. 1.2 x 10-5 J
d. 3.0 x 10-3 J
e. 0 J
The energy stored in the inductor is based on the current flowing through it. Since the
inductor and R1 are in the same branch, finding the current through R1 will do the trick.
Using Kirchoff’s rules, we can take the outer loop. After a long time, the voltage across
the inductor goes to zero which means the full all 9.0 V are dropped across R1. This
produces a 3 mA current in the branch with the inductor.
1
1
𝑈𝐿 = 𝐿𝐼 2 = (7.7 𝑚𝐻)(3 𝑚𝐴)2 = 3.5 × 10−8 𝐽
2
2
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PHYS 221
Exam 2
10 July 2015
3. A 24 V DC battery is connected to the primary of a transformer as shown. The transformer has
N1 = 5 turns and the transformer secondary has N2 = 10 turns. Assuming a that the magnetic flux
linkage is 100%, what is the voltage across R?
a. 0 V
b. 6 V
c. 48 V
d. 12 V
Transformers rely on Faraday’s law of induction to operate. Faraday’s law of
induction requires a magnetic field that changes over time, but a DC voltage
source creates a steady current which makes a constant magnetic field. There
is no induction, so there is no voltage across R.
4. The series RLC circuit shown below has R = 10 Ω, L = 1.0 H, and C = 1.0 μF. The AC generator is
applying a peak voltage given by ℰ = 14.14 V. What is the RMS voltage across the inductor at
resonance?
a. 0 V
b. 160 V
c. 225 V
d. 1000 V
e. 1414 V
The RMS voltage across the inductor is simple this: Δ𝑉𝐿,𝑅𝑀𝑆 = 𝐼𝑅𝑀𝑆 𝑋𝐿 .
As before, we are at resonance, so that means the impedance is just the resistance giving a peak current of
1.414 A. However, we need the RMS current, so we convert and obtain a 1 A.
Next we need the inductive reactance which is: 𝑋𝐿 = 𝜔𝐿 = (
that 𝜔 =
1
√𝐿𝐶
1
√𝐿𝐶
𝐿
) 𝐿 = √ = 1000Ω where we used the fact
𝐶
at resonance.
The final result is that the RMS voltage is 1000 V. This is greater than the voltage source, but remember that the
voltage across the capacitor cancels this out since they are equal, but out of phase by 180 degrees.
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PHYS 221
Exam 2
10 July 2015
5. Each chamber in the figure below has a unique magnetic field. A particle with charge q = 25 mC
and a mass of 10-10 kg enters the left chamber where B = 1.0 T directed into the page with a
velocity of 75 m/s. If the magnetic field in the second chamber is 0.5 T directed out of the page,
at what velocity does the particle leave chamber 2?
a. 37.5 m/s
b. 75 m/s
c. 150 m/s
d. 5625 m/s
e. Insufficient information to determine
Magnetic forces do no work so the incoming speed is the
outgoing speed.
6. Two charged particles are traveling in circular orbits with the same speed in a region of uniform
magnetic field that is directed into the page. The magnitude of the charge on each charge is
identical, but the signs are unequal.
𝑣
𝑣
𝐹
⃑
𝑣×𝐵
⃑
𝑣×𝐵
𝐹
Which one of the entries in the table is correct?
a.
b.
c.
d.
e.
Mass Relationship
m1 = m2
m1 > m2
m1 < m2
m1 > m2
m1 < m2
Sign of Q1
+
+
+
Sign of Q2
+
+
-
The relative direction of the cross product and the force gives us the sign of the charge. For the first particle they are
opposite to each other meaning the charge is negative. The second particle is then a positive charge.
If we use Newton’s second law with the understanding that we are undergoing uniform circular motion, then we get
the following equation: 𝑞𝑣𝐵 = 𝑚
first particle has a larger mass.
𝑣2
𝑟
5
. Solving for r shows us that the radius is proportional to the mass meansing the
PHYS 221
Exam 2
10 July 2015
7. A negative ion is moving east near the equator where the Earth’s magnetic field is horizontal to
the Earth’s surface pointing north. The direction of the magnetic force on the ion is:
⃑
𝐵
a. North
b. South
𝑣
c. Up
d. Down
e. There is no force
Marking out the velocity and magnetic field vectors makes it easier to use the right hand rule. Notice how to
rotate the velocity into the direction of the magnetic field we make a CCW rotation. This means that the cross
product is into the page.
However, the charge is negative. This means the force is in the opposite direction… i.e. it is up.
8. A positively charged particle moves through a magnetic field with a velocity that makes an angle
of 32° with the magnetic field. Describe the kind of motion the particle undergoes.
a. Parabolic
b. Uniform circular motion
c. Spiral
d. Straight line
e. Helical
Essentially a uniform magnetic field causes the component of the velocity that is perpendicular to it to
rotate. In a uniform field, this causes uniform circular motion perpendicular to the magnetic field. So if
the entire velocity is along the magnetic field, we just have a straight line… it it’s completely
perpendicular then we have uniform circular motion.
Anything in between is a combination of the two… helical motion.
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