Physics Exam 3

b (a) As the capacitor charges, the current decreases from its initial maximum value to zero. (b) Therefore, the lightbulb begins at its brightest and dims with time.

A capacitor can be charged by the battery when the switch at the right is closed. If the capacitor is initially uncharged, how does the brightness of the bulb change with time after the switch is closed? (a) The bulb becomes brighter with time. (b) The bulb becomes dimmer with time. (c) The bulb glows with a constant brightness.

b The potential difference across the capacitor decays as a function of time.

A capacitor is charged to potential difference V and then allowed to discharge through a resistor. How does the potential difference across the capacitor change with time as the capacitor discharges? (a) The capacitor potential difference increases with time. (b) The capacitor potential difference decreases with time. (c) The capacitor potential difference stays the same.

b The energy stored in an inductor is U = (1/2) LI^2 = (1/2) (0.003H)(0.5A)^2 = 4 × 10^−4 J

A commercial inductor has inductance 0.003H and carries current 0.5A. How much energy is stored in the magnetic field of the inductor? (a) 8 × 10^−4J (b) 4 × 10^−4J (c) 2 × 10^−3J (d) 1 × 10^−6J (e) 2 × 10^−6J

g Zero. The interaction of a magnetic field and a charged object moving with respect to the field produces a Lorentz force perpendicular to both the field and the object's velocity. If the charged object has no velocity with respect to the field, there is no Lorentz force.

A non-uniform magnetic field points in the +ˆy direction at the origin. The field increases in strength as y increases. What is the direction magnetic force on a positively charged object (with zero magnetic moment) stationary at the origin? (a) +ˆx (b) −xˆ (c) +ˆy (d) −yˆ (e) +ˆz (f) −zˆ (g) The force is zero.

g Zero. Every moving charge in the universe produces a magnetic field, but an unmoving or stationary charge does not produce a magnetic field.

A positive charge is fixed at the origin. What is the direction of the magnetic field at a point P at +1cmˆy? (a) +x (b) −x (c) +y (d) −y (e) +z (f) −z (g) zero

b Changing the particles sign would change the direction of rotation.

A positively charged particle orbits in a clockwise direction in the field to the right. If the particle were to suddenly become negatively charged, what would happen to the orbit? (a) The orbit would become a spiral. (b) The orbit would change direction; the particle would travel in a counterclockwise direction. (c) The trajectory would become a straight line. (d) Nothing would happen to the orbit. (e) The particle could not move through the field if it was negative.

b The magnetic energy stored in a system of conductors is U = (1/2) LI^2 where L is the inductance. U = (1/2) LI^2 = (1/2) (9.7 × 10^−7H)(100, 000A)^2 = 4.85 × 10^3 J

A rail gun has inductance 9.7 × 10−7H when the projectile leaves the rail. The peak current is 100, 000A. Compute the energy stored in the rail gun at peak current. (a) 9.7 × 10^3J (b) 4.85 × 10^3J (c) 2.2 × 10^−5J (d) 9.7 × 10^−2J

c The time constant for an LR circuit is L/R = 0.02H/100Ω = 0.0002s

A resistor with resistance 100Ω is connected in series to an inductor with inductance 0.02H and a battery at time t = 0. How long does it take the current through the series combination to reach 1 − e −1 = 0.63 of its long time value? (a) 2s. (b) 0.02s. (c) 0.0002s. (d) 200s

b The initial voltage of the resistor in an RC circuit is the full battery voltage.

A resistor with resistance R, battery with voltage ∆V , capacitor with capacitance C, and switch are connected in series. The switch is closed at t = 0. The time constant of the circuit is τ. What is the initial voltage across the resistor immediately after the student is closed? (a) 0 (b) ∆V (c) τ∆V (d) ∆V /τ (e) ∆V /(R + C)

d The strength of the field is inversely proportional to r^2 , so the field is one-quarter the strength at B.

A small element of current I∆ℓ runs through the origin in the direction drawn. The element is part of a larger current carrying wire. Compare the strengths of the magnetic field BA at a point A and the magnetic field BB at point B due to the element. The point B is twice as far from the element as point A. (a) BA = BB = 0 (b) BA = BB 6= 0 (c) BA = 2BB (d) BA = 4BB (e) BA = BB/2 (f) BA = BB/4

a The magnitude of the magnetic field B inside of a solenoid with length l and N wrappings with current I in each wrapping is B = µoIN/l . If the solenoid length and number of wrappings are both doubled, the solenoid's magnetic field would remain unchanged.

A solenoid in the infinite solenoid approximation is wound with n turns per unit length. The solenoid has length L and carries current I producing a magnetic field B. If the length of the solenoid is doubled while the current and the turns per unit length stay the same, what is the new magnetic field in the solenoid? (a) B (b) B/2 (c) B/4 (d) 2B (e) 4B

b The self-inductance of a solenoid increases linearly with the area, so the larger solenoid has twice the inductance.

A solenoid with length ℓ, N turns, and cross-sectional area A has inductance L if the solenoid can be treated as infinitely long. What would the inductance become if the area was doubled while the other parameters remained the same? (a) L (b) 2L (c) 4L (d) L/2 (e) L/4

b The field out of the page is decreasing and therefore the flux out of the page is decreasing; the induced current will produce a flux out of the page by Lenz' law. By the right hand rule, that requires a counterclockwise current.

A square coil of wire sits in a magnetic field directed as shown. The magnetic field is decreasing in strength. The direction of the induced current is: (a) clockwise (b) counterclockwise (c) The current is zero.

b The emf across an inductor is given by Faraday's law, emf = − dφm/dt = −L dI(t)/dt = −5Lγt^4

A time varying current is applied to a commercial inductor with inductance L. The current varies as a function of time according to I(t) = γt5 where γ is a constant. The emf across the inductor as a function of time is: (a) −Lγt^5 . (b) −5Lγt^4 . (c) −(Lγt^4)/5. (d) −5Lγt^6 . (e) Zero.

a Current is net flow of charge. If equal charge is flowing in opposite directions, then zero net charge is flowing.

A two-conductor cable is connected at one end so that a current I flows through the conductors. The current flows up one conductor and back down the other. A segment of the cable is drawn to the right. What is the net current flowing through a ring whose boundary encircles both conductors? (a) 0 (b) I (c) 2I (d) 4I (e) I/2

d φm = LI = (0.022H)(0.05A) = 1.1 × 10^−3 Tm^2

According to information found on the Internet years ago, the Series 4922 Power Inductor from Delevan has a maximum current rating of 0.050A for the 0.022H inductor. At maximum current, what is the magnetic flux through the inductor? (a) 25Tm^2 (b) 5.3Tm^2 (c) 4.4 × 10^−1 Tm^2 (d) 1.1 × 10^−3 Tm^2 (e) 3.1 × 10^−5 Tm^2

c An infinite solenoid produces a uniform magnetic field in its interior.

For some application, you are asked to produce a uniform magnetic field (in some region) using equipment you already have in the lab. Which of the following produces a uniform field? (a) disk shaped ceramic magnet (b) infinite straight wire (c) infinite solenoid (d) single loop of wire

d 10% of ∆Vo is 1V . The plot reaches 1V at t = 22.5s.

Given the plot below of the potential difference across a resistor through which a capacitor is discharging: How long does it take for ∆VR to decay to 10 % of its value at t = 0? (a) 6s (b) 10s (c) 15.5s (d) 22.5s (e) 26.5s

b For example, if a stationary charge is present in some region of space, an electric field is present without a magnetic field.

In a region of space, there exists a non-zero electric field. What can be deduced about the magnetic field in the region? (a) There must be a magnetic field in the region. (b) There may or may not be a magnetic field in the region. (c) There cannot be a magnetic field in the region.

a The energy of a system of two point charges is the work to build the system. U = W = qi∆V2 = kq1q2/d Now, substitute the values given in the problem. U = (8.99 × 10*9 Nm^2/C^2 )(40C)(40C)/1000m = 1.4 × 10^10J Not bad.

In a thundercloud, separated regions of charge with charge 40C can develop. If the regions of charge are 1000m apart, how much energy is stored in the separated charge. You may model the charges as point charges. (a) 1.44 × 10^10J (b) 1.42 × 10^−11J (c) 3.60 × 10^8J (d) 1.42 × 10^−14J (e) 1.44 × 10^7J

d The Lorentz Force. As charge flows through the wire, the magnetic field applies a force (Lorentz Force) on the moving charge. Because the force is perpendicular to both the direction of the field and the direction of the current, the rod deflects.

In lab, you performed an experiment on something I called a "magnetic pith ball" where an aluminum rod was suspended by copper wire so it hung in one of the big lab magnets. When a current was run through the circuit, the rod deflected. This happened because of which of the following physical laws? (a) Gauss' Law (b) Ampere's Law (c) Faraday's Law (d) Lorentz force (e) Heisenburg Uncertainty Principle

a A circuit with two capacitors in series is shown to the right.

Is the combination of capacitors at the right a series or parallel combination? (a) series (b) parallel

d For a discharging RC circuit, the time dependence of the current is I(t) = Ioe^−t/τ where τ = RC = 4 × 10^−6 s is the time constant. Solve for t, t = −τ ln (I/Io) = −(4 × 10^−6 s) ln ( 0.05 )= 12 × 10^−6 s

Our big lab capacitor has a capacitance of 2000µF. If the capacitor is discharged through a copper wire (resistance 0.002Ω), how long does it take for current delivered by the capacitor to drop to 5% of its initial value? (I = 0.05I0) (a) 3.8 × 10^−6 s (b) 4.6 × 10^−6 s (c) 2.1 × 10^−7 s (d) 1.2 × 10^−5 s

a The charge on the capacitors is the same as the charge on the equivalent capacitance.

Problem: Two capacitors, each with capacitance C, are connected in series and have equivalent capacitance Ceq . The combination is connected to a battery. How is the charge Qeq on one plate of the equivalent capacitor related to the charge, Q, on one plate of the individual capacitors? (a) Qeq = Q (b) Qeq = Q/2 (c) Qeq = 2Q (d) Qeq = Q/4 (e) Qeq = 4Q

f The cross product 3^x × 2ˆz can be evaluated using the Right Hand Rule. Point your fingers in the direction of A ⇒ x^, then curl your fingers in the direction of B ⇒ zˆ; your thumb should be pointing in the −yˆ direction. Since A and B are perpendicular, simply multiply the magnitudes to get the size of the resulting vector, |A~||B~ |sin 90◦ = |A||B|. The cross product is 3ˆx × 2ˆz = −6ˆy.

What is the cross product, A × B, of the vectors A= 3ˆx and B= 2ˆz? (a) zˆ (b) xˆ (c) −6ˆx (d) +6ˆx (e) +6ˆy (f) −6ˆy (g) +6ˆz (h) −6ˆz

d

What is the name of the central physical law explaining the behavior of a generator? (a) Lorentz Force (b) Gauss' Law (c) Ampere's Law (d) Faraday's Law

c The magnetic force acts perpendicular to field lines. Mathematically, the direction of the magnetic force is found by the cross product between charge-velocity and magnetic field (F~ = q~v × B~ ). Because the force is the result of a cross product involving the magnetic field, it is always perpendicular to the field.

What is the relation between the direction of the magnetic field and the magnetic force? (a) The magnetic force and field are parallel. (b) The magnetic force and field point in opposite directions. (c) The magnetic force and field are perpendicular. (d) There is no general relation between the direction of the magnetic force and field.

d The needle on a compass is a small bar magnet,

What physically is a compass needle? (a) a dielectric with a net charge (b) a conductor with a net charge (c) a special material that naturally finds north (d) a permanent magnet (e) a small piece of copper

d

Which of Maxwell's equations is most useful in the calculation of the magnetic fields of static current distributions? (a) Gauss' Law (b) No Magnetic Monopoles (c) Faraday's Law (d) Ampere's Law

c

Which of Maxwell's equations is of primary importance in the generation of electrical power? (a) Gauss' Law (b) No Magnetic Monopoles (c) Faraday's Law (d) Ampere's Law

f By the right-hand-rule, if x points to the right and z points inward, y must point toward the bottom of the page (point your fingers to the right, your thumb inward, and curl your fingers; they will point downward).

You draw a coordinate system with +x pointing to the right of the page and +z pointing into the page. In what direction is +y? (a) into the page (b) out of the page (c) to the left of the page (d) to the right of the page (e) to the top of the page (f) to the bottom of the page

a The time constant is proportional to the resistance. The resistance of circuit (a) and (b) are equal and larger than that of circuit (c).

You have a capacitor and two resistors. How should the resistors be connected to produce the circuit with the highest time constant? (a) series (b) parallel (c) Either series or parallel connections will yield the same time constant.

g The magnetic force by field B on a particle with charge q and velocity v is given by F = qv × B. Since the charge is stationary, the magnetic force is zero

An electron is at rest a point P. A current I flows in the wire in the direction drawn. What is the direction of the magnetic force on the electron. Electrons are negatively charged. (a) +ˆx (b) +ˆy (c) +ˆz (d) −xˆ (e) −yˆ (f) −zˆ (g) zero

e The magnetic field outside an infinite solenoid is 0.

An infinite solenoid is wound with 100 turns per meter and carries a current of 8A. The solenoid is cylindrical with radius 10cm. What is the magnetic field OUTSIDE the solenoid at a distance of 20cm from the axis? (a) 1.0 × 10^−3T (b) 3.0 × 10^−5T (c) 1.5 × 10^−5T (d) 1.0 × 10^−6T (e) 0

d Using the current directions given in the figure above, if we conserve current at node j, we find: I2 = I1 + I3 Current into the junction is positive, and current out of the junction is negative for sum of I initial = 0.

Analyze the circuit to the right with ∆V1 = 12V , ∆V2 = 6V , R1 = R2 = R3 = 10Ω. Use the directions for the currents drawn on the diagram. Write a junction equation for junction j. (a) I2 = I1 (b) I1 = I2 + I3 (c) I3 = I1 + I2 (d) I2 = I1 + I3 (e) 0 = I1 + I2 + I3

e Using the right-hand rule for the magnetic field due to a current-carrying wire, the magnetic field at the center point due to a counterclockwise current is pointing upward.

The circular loop of wire to the right lies in the x − y plane and carries current in the counterclockwise direction as viewed from a point on the positive z axis as drawn. What is the direction of the magnetic field at the point P along the positive z axis? (a) into the page (b) out of the page (c) to the left of the page (d) to the right of the page (e) to the top of the page (f) to the bottom of the page (g) The field is zero.

b The field of the bottom wire at the top wire is B = µoI/2πd = (4π × 10^−7 Tm/A )(17A)/2π(0.006m) = 5.667 × 10^−4T The Lorenz force is F = IL × B. Since the current and magnetic field are at right angles, F = ILB = (17A)(0.265m)(5.667 × 10^−4T) = 0.0026N

The current balance wires are 26.5cm long. At maximum current, the device carries a current of 17A. The normal equilibrium center to center separation of the wires is 0.60cm. Compute the force on the top wire due to the magnetic field of the bottom wire. (Treat the bottom wire as an infinite straight wire.) (a) 0.0096N (b) 0.0026N (c) 0 (d) 14N (e) 0.43N

a The current element points in the direction of the current, x.

The drawing to the right shows a wire carrying current I in the direction indicated. What is the direction of the current at point A? (a) +x (b) −x (c) +y (d) −y

b First, convert bulbs and batteries to their circuit representation. Next, are the bulbs in series or parallel? If one bulb was unscrewed, the other bulb would remain lit because there is still a potential difference. Therefore, the circuit is in parallel.

The figure below and to the left shows the physical representation of a battery connected to two light bulbs. The light bulbs are shown in sockets that provide electrical connection points. Select the schematic representation of the same circuit. (a) Figure (a) (b) Figure (b) (c) Figure (c) (d) Figure (d)

B The junction rule for currents states that the current going into a junction must equal the current going out of a junction. Using the current directions given in the figure above, if we conserve current at node b, we find: I1 = I2 + I3.

The figure below shows a circuit that is part of a larger circuit. Select the equation below that is the junction equation for junction c using the current directions drawn. (a) I2 + I3 = 0 (b) I1 = I2 + I3 (c) I1 + I2 = I3 (d) I1 = I3 (e) I1 + I2 + I3 = 0

b To oppose the increasing flux caused by the increasing field into the page, a magnetic field out of the page will be created by the induced current. Using the right hand rule for a wire, this gives the direction of induced current flow as counterclockwise.

The figure to the right shows a magnetic field whose strength is increasing with time and whose direction is into the page. Is a current induced in the conducting ring? If so, what is the direction of the induced current in the conducting ring (shown as a dark circle)? (a) clockwise (b) counterclockwise (c) There is no induced current.

d Applying Ohm's law to the top resistor allows the calculation of the potential difference between a and b, |∆Vab| = IR = (2A)(4Ω) = 8V This is also the potential difference across the bottom resistor, because they are in parallel. The current through the bottom resistor is found by applying Ohm's law, I2 = |∆Vab|/R = 8V/2Ω = 4A

The figure to the right shows a parallel combination of resistors that is part of a larger circuit. Current I1 = 2A flows through the top resistor. The top resistor is 4Ω and the bottom resistor is 2Ω. Calculate I2, the current through the bottom resistor. (a) 1/2A (b) 1A (c) 2A (d) 4A (e) 8A

a (a) In order for the magnetic field to be zero at point a an equal and opposite magnetic field must be created in the wire to oppose the leftward uniform magnetic field. (b) Current flowing into the page would create a magnetic at point P that points to the right that would cancel out the uniform magnetic field to the left.

The figure to the right shows a uniform magnetic field pointing to the left of the page. An infinite straight wire passes through the point drawn. In what direction must current flow in the wire to produce zero magnetic field at the point P? (a) into the page (b) out of the page (c) to the left of the page (d) to the right of the page (e) to the top of the page (f) to the bottom of the page (g) No direction of current in the wire will produce zero total field at P.

e- Before the switch is closed, there are four resistors in series for an equivalent resistance of 4R.If the switch is closed, all current goes through the switch and none through resistors R2 and R3.Therefore, these two resistors can be removed from the circuit.This leaves two resistors in series R1 = R2 for an equivalent resistance of Req = 2R. Since the resistance is halved, the current is doubled: ∆Vbatt = IReq.

The figure to the right shows an arrangement of identical resistors in a circuit. With the switch open, a current I flows in the circuit. How much current flows in the circuit if the switch is closed? (a) 0 (b) I (c) I/2 (d) I/4 (e) 2I (f) 4I

e The currrent enclosed is the signed sum of the currents passing through the surface. Let out of the page be positive current. Path 1 encircles 2I, Path 2 encircles 2I − I = I, and Path 3 zero

The figure to the right shows three wires either carrying current into or out of the page. The wires on the outside carry current I and the central wire a current of 2I. Three Amperian paths are drawn; rank the current encircled, I1, I2, and I3 for the three paths. Assume a current out of the page is positive. (a) I1 = I2 = I3 (b) I1 > I2 = I3 (c) I2 > I1 > I3 (d) I1 < I2 < I3 (e) I1 > I2 > I3

d The mutual inductance is defined as MAB = φB/IA = 15Wb/3A = 5H

Two identical coils are positioned as shown at the right. The side view of each coil is shown. When a current IA = 3A is passed through coil A, a magnetic flux of 25Wb is produced through coil A and a magnetic flux of 15Wb is produced through coil B. Compute the mutual inductance MAB. (a) 75H (b) 45H (c) 25/3H (d) 5H (e) 5/3H