Solutions for Exam #2 Prepared by Kurt Jung Questions are written

Solutions for Exam #2
Prepared by Kurt Jung
Questions are written out so if you had a different test you can figure out which one I’m
talking about. Numbers in bold will change from test to test, but I solved with the
numbers given. Procedures should work the same way.
1. For the circuit show above, E = 10 V and R = 7 ohms. Find the potential
difference between points a and b. State the magnitude.
This one is a difficult problem, but in the end, we need to solve for the current in the
middle segment of the circuit so we can find the voltage drop across R. We can use
Kirchhoff’s loop rules and node rules to find a system of equations such that we can solve
for the middle segment’s current (which I called i3). So from the left side (taking current
counterclockwise), I have:
From the right side loop I have:
These equations are exactly the same except for i1 and i2, therefore, i1 = i2, and, by the
node rule, i1+i2 = i3, or 2 i1 = 2 i2 = i3. Substituting this in with the numbers given
allows us to solve for i3 = 1 A. Therefore, the voltage drop across R is 7 V (V = i*R), so
the voltage difference between points a and b must be 3 V.
2. A certain dielectric with a dielectric constant K = 24 can withstand an electric
field of 4 x 107 V/m. Suppose we want to use this dielectric to construct a 0.87 uF
capacitor that can withstand a potential difference of 2360 V. (a) What is the
minimum plate separation? (b) What must the area of the plates be?
For this one, notice that:
So since E is constant between the plates of a parallel plate capacitor, l = V/E = 6.6 x 10-5
m, or about 0.06 mm. Then, we know that for a parallel plate capacitor
So since we know d, k, e0 and C, we can rearrange to solve for A, which gives us 2416
3. In the figure above, the emf is 5.9 V and R = 0.54 ohms. The rate of Joule
heating in R is 8 W. What is the value of r?
“The rate of joule heating” is a fancy way of talking about the dissipated power in the
resistor, or P = I2 R. Rearranging gives me that I = 3.85 A, but since there’s only one
loop, the current through the resistor must be the same everywhere. Using the loop rule,
V – i r – i R = 0, or r = 0.99 ohms.
4. When a current is passed through the flexible wire shown in the Figure below,
will it tend to bunch up or form a circle?
Magnetic field is generated from a current carrying wire, circulating around the wire by
the right hand rule. Then, the force on a current carrying wire is expressed as
So using the right hand rule again gives that the forces always repel from wires carrying
opposite currents, next to each other, as is the case in the figure. Therefore, the wire
tends to form a circle.
5. A solenoid 2.9 m long has a radius of 0.85 cm and 680 turns. It carries a current I
of 2.5 A. What is the approximate magnetic field B on the axis of the solenoid?
For solenoids, I know that
And that
So setting these equal gives me that flux = N B A = u0 n2 A l I. Rearranging and solving
gives that B should be 0.737 mT.
6. What is the maximum torque on a 900 turn circular coil of radius 0.75 cm that
carries a current of 2.2 mA and resides in a uniform magnetic field of 0.25 T?
We know that torque on a current carrying loop is expressed as:
Where u (the magnetic moment) is expressed as
The maximum torque will occur when the angle between u and B is 90 degrees, or when
torque is simply u x B. Therefore, max torque = N I A B = 8.75 x 10-5 N m
7. The circuit as shown in the Figure above shows a capacitor C, two ideal batteries,
two resistors, and a switch S. Initially S has been open for a long time. If it is
then closed for a long time, by how much does the charge on the capacitor
change? Assume C = 72 uF, e1 = 2.2 V, e2 = 0.77 V, R1 = 0.14 ohms, R2 = 0.62
I thought this was the hardest question on the test and had some trouble coming up with
an answer, but I’m pretty sure this is correct. So when the switch is open, notice that we
have a simple circuit with a battery, capacitor, and resistor. Therefore, at time t=infinity,
the capacitor is fully charged, hence the current is 0 in the circuit. So the battery must
have a voltage equal to that of the battery. Therefore,
Q(initial) = C / e2 = 5.544 x 10-5 C.
Now notice that when the switch is closed, we have three complete loops, but after a long
time, the capacitor will become fully charged and no current will run through the middle
of the circuit. Therefore, what we must do is find the total current in the outside loop
using Kirchhoff’s loop rule. Moving counterclockwise, we have
e1 – i R1 – i R2 – e2 = 0
Therefore (using my numbers) the current in the circuit is 1.88 A. Using this, we can use
Kirchhoff’s loop rule again on the left (or right) loop to find the voltage drop across the
capacitor. So using the left loop, we see that
E1 – i R1 – V(cap) = 0.
Therefore, V(cap) = 1.937 V. Then, finally, our
Q(final) = C / V(cap) = 1.397 x 10-4 C.
So the change in charge is Q(final) – Q(initial) = 84.2 x 10-6 C ~ 84 uC.
8. An electric field of 1.8 kV/m and a magnetic field of 0.56 T act on a moving
electron to produce no net force. Calculate the minimum speed V of the electron.
Since there’s no net force, we know that Fe = Fb or qE = qv x B. The minimum speed
required is when v and B are perpendicular, so we have that qE = qvB or v = E/B.
Plugging in numbers gives v(min) = 3.21 x 103 m/s
9. In the circuit shown in the Figure above, e = 10 V, R1 = 3.9 ohms, R2 = 9.2 ohms,
and L = 5.4 H. for the situation where the switch S just closed, calculate the
current i1 through R1.
Remember that as the switch is just closed, the change of current is very high. Therefore,
the voltage across the inductor is very large, so almost no current flows through that right
side of the circuit. Therefore, since (almost) all of our current produced by the battery
will be i1, it’s a simple matter of V = IR to calculate the current. So i1 = e/R1 = 2.56 A.
10. A 10-gauge copper wire (2.6 mm in diameter) can carry a current of 50 A without
overheating. For this current, what is the magnetic field at the surface of the
This problem is a simple application of Ampere’s Law. Notice that since we want the
magnetic field just at the surface of the wire, we can draw an Amperian loop just around
the wire, thereby enclosing all the current. Ampere’s Law is:
∫ B⋅ dl = µ I
0 encl
Therefore, since B is independent of the position of the loop, we pull the B through the
integral sign and rearrange to get
B = 0 encl
Finally, since we enclose all the current with our amperian loop, we simply plug in and
solve. With my numbers, I get that B = 7.69 x 10-3 T = 7.69 mT.
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