Midterm #1 - Department of Physics
Exam 2 Solutions
1. An RC circuit is discharged by closing a switch at time t = 0. The initial potential difference across the capacitor is 5 V. The potential difference across the capacitor drops to half of its value in 35 ms.
a) [6 points] What is the time constant of the circuit?
The expression for the potential difference across the capacitor when discharging is given by [pic]
If the voltage drops to half of its value,
[pic]
b) [4 points] If the total resistance in the circuit is 10 k(, what is the capacitance?
[pic]
2. [8 points] In the circuit shown, the resistances are[pic], [pic], and [pic]. The battery voltages are [pic]and [pic]. What is the current (in amps) flowing through the battery with potential difference[pic]?
[pic]
Let [pic] be the current through the (1, 18V, battery, and [pic] the current through the right branch with the (2, 6V, battery, and [pic] the current through the middle branch.
By Kirchoff’s junction rule: [pic].
By Kirchoff’s voltage sum rule applied to the rightmost loop:
[pic]
By Kirchoff’s voltage sum rule applied to the outer loop:
[pic]
Answer: 1.0 Amps through 18V battery
3. [6 points] Two identical amounts (volumes) of copper are formed into two cylindrical rods, the first of length L and the second of length L/2. If the resistances of the two volumes (measured between the two ends) are [pic] and [pic] respectively, what is [pic]?
Resistance is given by the expression:
[pic]
The key point is that the volume remains the same, so the cross sectional area must change.
Thus
[pic]
4. [6 points] A large flat non-conducting sheet has an electric charge density of [pic], where [pic], as measured in the rest frame of the sheet. What is the charge density “seen” by an electron traveling at a velocity of [pic] parallel to the surface of the sheet?
Length is contracted along the direction of motion: [pic], but not in directions perpendicular to the direction of motion. However, the electric charge in some region (let’s say a square in the rest frame of the sheet), must remain invariant between different reference frames. So if the square area on the sheet gets contracted along the direction of motion (becoming a rectangle as observed by the electron), the area decreases while the charge remains the same. So the charge density should increase.
In the rest frame of the sheet, the area of the square is [pic] and the charge is [pic], where (0 is the charge density in the rest frame.
In the frame of the electron, the sheet area is contracted to: [pic]
The observed charge density is thus:
[pic]
That is, the charge density appears higher to the electron by the factor (. Plugging in:
[pic]
5. [6 points] Galaxy A is reported to be receding from us at a speed of 0.41c. Galaxy B, located in precisely the opposite direction, is found to be receding from us at the same speed. What recessional speed would an observer on Galaxy A find for Galaxy B?
The velocity of Galaxy B in the rest frame of us on Earth is 0.41c. We want to transform this velocity to a frame moving at a velocity of -0.41c moving in the opposite direction.
Use the Lorentz transformation of velocity:
[pic]
Note that this is not the Galilean expectation of 0.82c.
6. [6 points] An electron is accelerated from rest by a potential difference of 10,000 V. It then enters a uniform magnetic field of magnitude 0.05 T with its velocity perpendicular to the direction of the field. Calculate the radius in meters of its path in the magnetic field. The mass of the electron is [pic].
Setting the centripetal acceleration, [pic], equal to the acceleration due to the magnetic force, [pic], yields the following relation for the radius of curvature for a charged particle moving in a uniform magnetic field:
[pic]
To get the velocity, one must realize that the kinetic energy of the electron after accelerating from rest across an electric potential equals the initial potential energy:
[pic]
Answer: [pic]
7. The rectangular-shaped loop shown in the figure is carrying a current i = 2A in the clockwise direction shown. The length of the vertical sides is 7cm and the length of the horizontal sides is 3.5 cm. A uniform magnitude field B = 0.05T makes an angle (=30( with respect to the plane of the plane of the loop.
(a) [6 points] What is the magnitude of the torque acting on the loop
if the left side is connected to a hinge?
[pic]
Note, by the right-hand rule, the magnetic dipole moment [pic] points in the [pic] direction. The torque is given by:
[pic]
(b) [6 points] If the loop is free to rotate about its hinge due to the torque, by how much will the potential energy of the loop change when the torque becomes zero?
Potential energy of a dipole in a magnetic field is given by:
[pic]
8. [8 points] A long straight wire carries a current [pic] in the horizontal direction shown. Below it is a square loop of side length [pic] carrying a current [pic] in the clockwise direction shown. The distance between the top of loop to the long straight wire is distance a, which you can take to be much smaller than b ([pic]). If the mass of the loop is [pic]10 g, what is the maximum distance a such that the loop is attracted to the wire with current [pic] with a force that overcomes the force of gravity? (the acceleration due to gravity points in the [pic]direction).
[pic]
The magnitude of the force between two parallel current-carrying wires is:
[pic]
The force on the top wire is attractive (up) since the currents are in the same direction, and the force on the bottom wire is repulsive (down). By symmetry, the force on the two vertical sides will cancel (and are actually zero because [pic]).
The net force is up since the distance to the bottom wire is larger:
[pic]
9. [6 points] A horizontal power line carries a current of 3000A from south to north. Earth's magnetic field, with a magnitude of 60(T, is directed toward the north with a dip angle 60° downward into the Earth relative to the horizontal. Find the magnitude and direction (use compass directions) of the magnetic force acting on a 100 m length of power line.
The magnetic field and the power line both point north, but the magnetic field points into the Earth at an angle of 60°.
[pic]
The force on a current carrying wire is given by:
[pic]
So the force points west with magnitude:
[pic]
Answer: 15.6 N
10. [8 points] An infinitely long insulated wire carrying a current I = 100 A is bent into the shape shown (straight line plus circle of radius R=3 cm with the currents in the direction shown). Find the magnitude of the field B at the center of the circle.
[pic]
We calculate the field from both the long wire and the circle separately. From the long wire, the field at the center of the circle, a distance R away, is:
[pic] into the plane of the paper.
From the circular wire, the field is:
[pic]
Since the magnetic field from the arc is in the opposite direction of that from the long wire, the total magnitude is the difference:
[pic]
Answer: [pic]
11. [6 points] A conductor has a cylindrical cross section with a radius a = 2 mm. Inside, a cylindrical cut-out hole (non-conducting) is made as shown, where the radius of the cut-out is b = 0.5 mm and the distance between the center of the main conductor and the center of the cut-out is d = 1 mm. The magnitude of the current density in the conductor is[pic]. What is the magnitude of the magnetic field at a radius of [pic]mm from the center of the conductor along a line connecting to the center of the cut-out? [Hint: try using the superposition principle.]
[pic]
We break this problem up into two contributions: the magnetic field due to a solid conductor of radius a and with a current coming out of the page, and the magnetic field from a smaller conductor of radius b with a current going into the page. (The two currents will cancel, leading to an effective cavity in the shown conductor).
By Ampere’s Law, the field inside the large cylindrical conductor at a radius r can be obtained from:
[pic]
It points in the counter-clockwise direction for a current out of the page and for r=0.5 mm to the right of the center.
The field from the smaller conductor has the exact same magnitude since the current density is equal and opposite. Also, the radius r=0.5mm places the point for the field to be calculated at the same radius from the center of conductor b. Thus:
[pic]
It points in the clockwise direction, but at r=0.5mm from the center of conductor a it points in the same direction! So the two fields add, and we have:
[pic]
12. A rectangular loop of wire of sides [pic] is moving with constant speed v into a region of uniform magnetic field of magnitude B directed out of the paper as shown in the figure (the long side is along the direction of motion).
a) [6 points] What is the magnitude of the induced EMF in the
wire and the direction of the induced current just
after the loop enters the region of magnetic field?
[pic]
Faraday’s Law: [pic]
The magnetic field is a constant, but the area in the field region changes:
[pic], so
[pic]
By Lenz’s Law, the current is in the direction such that it creates a magnetic field that opposes the changing flux. Since the flux is increasing through the loop out of the page, we need to add a field into the page which can be accomplished with a clockwise current.
b) [6 points] If the total resistance of the wire is R, what is the power dissipated by the loop?
[pic]
13. [6 points] A circular loop of wire of radius 30 cm sits in the x-y plane in a region where there is a uniform magnetic field of [pic], where B0 = 0.5 T. The magnetic field is now increased at a constant rate such that it reaches three times its original strength in 20 s. What is the magnitude of the induced EMF in the wire loop during this time?
Faraday’s Law: [pic]
The loop area remains the same, only the magnetic field changes:
[pic]
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PHY2061
11-8-05
Name:_______________________ ___
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