# Exam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses.

Size: px
Start display at page:

Download "Exam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses."

Transcription

1 Exam 2 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Part of a long, straight insulated wire carrying current i is bent into a circular section of radius d as shown in the figure. What is the magnitude of the magnetic field at the center of circular section if both straight and curved sections lie in the plane of the page as shown? (1) (μ0 i / 2 π d) (π - 1) (2) (μ0 i / 2 π d) (π + 1) (3) (μ0 i / 2 π d) (4) (μ0 i / 2 d) (5) 0 The field contribution from the straight wire at the center of the circular µ points into the page with magnitude 0 i 2πd The field contribution from the loop at its center points out of the page. You can use the Biot- Savart Law to compute it: db = µ 0i ds ˆr = µ 0i rdφ 4π r 2 4π r 2 B = µ 0i 2d So the magnitude is the difference: B tot = µ 0i 2πd π 1 ( )

2 Problem 2 A circular loop of radius 0.1m is oriented in the xy- plane and carries a current i=2a in the counter- clockwise direction as shown. It is allowed to rotate about the x- axis. A uniform magnetic field B of magnitude 0.3T is applied in the y direction. What is the magnitude of the torque (in N- m) about the x- axis, and in what direction does the top part of the shown loop move? (1) 0.02, into page (2) 0.02, out of page (3) 0.06, into page (4) 0.06, out of page (5) 0, does not move µ = ia points out page τ = µ B points in î direction τ = iπr 2 B = 0.02 N-m Since the torque is in the i direction, the top of the loop moves into the page (also can be worked out from df = idl B) Problem 3 An electron with a velocity vector of v = 10,000 (i+j)m/s moves through a uniform magnetic field given by B = 4 (- i + j)t. What is the force (in N) on the electron? (1) k (2) k (3) k (4) k (5) - 40,000 j F = qv B = 4( 10000)e î + ĵ ( ) ( î + ĵ ) ( ) = 40000e î ĵ ĵ î = 80000eˆk = ˆk

3 Problem 4 A power line that is horizontal to the surface of the Earth carries a current of 9000A from north to south. Earth's magnetic field of magnitude 60μT is directed toward the north and is inclined downward from horizontal at an angle of 60. What is the magnitude of the force per unit length on the power line and its direction? (1) 0.47 N/m, east (2) 0.47 N/m, west (3) 0.54 N/m, west (4) 0.54 N/m, east (5) 0.27 N/m, south S i 60 N B The force on a wire carrying a current in a magnetic field is given by: F = il B For the drawing shown, this is going to lead to a force coming out of the page. This would be in the eastern direction. The magnitude of the force per unit length is: F l = ( 9000 )( )sin120 = 0.47N Problem 5 In the circuit shown, ε1=3.0v and ε2=ε3=1.5v. Also R1=2 Ω and R2=R3=1 Ω. What is the magnitude of the current flowing through R2? (1) 0.3A (2) 1.5A (3) 0.6A (4) 0.9A (5) 0

4 Let each current point vertically up through each EMF source. Since ε2=ε3 and R2=R3, the current i2 = i3 (the branches are equivalent). By the junction rule we have i1 = - 2 i2 Then by the loop rule for the outermost loop: ε 1 R 1 i 1 + R 2 ε 2 = 0 ( ε 1 ε 2 ) + 2R 1 + R 2 = 0 plug in junction rule ( ε 1 ε 2 ) = 2R 1 R 2 ( ) = ε 1 ε 2 = 1.5 2R 1 + R 2 5 = 0.3 Problem 6 What is the equivalent resistance of the shown circuit as seen by the EMF source if all resistors have resistance R? (1) 3R/5 (2) R/4 (3) R (4) 4 R (5) 5R/2 The 2 resistors on the right are in parallel with equivalent resistance R/2. These are in series with R at the top, for resistance 3R/2. That in turn is in parallel with a resistance R, so the total equivalent resistance is 3R/5. Problem 7 The current density j inside a long, solid, cylindrical wire of radius a is in the direction of the central axis, and its magnitude varies inversely with radial distance r from the axis according to j = j0 a/r. Find the magnitude of the magnetic field at r=a/2. (1) μ0 j0 a (2) μ0 j0 a / 2 (3) μ0 j0 a / 3 (4) μ0 j0 a / 4 (5) μ0 j0 a ln(a/2)

5 Problem 8 Use Ampere's law. By symmetry, the magnetic field is directed axially around the wire, and its magnitude depends only on the distance from the central axis. We can best exploit this by choosing the closed path to be a circle of radius r, perpendicular to the central axis and centered on it. The boundary of such a circle is its circumference so ds B = 2πrB(r) By Ampere's law this must be µ 0 2π dr r j( r ) so B(r) = µ 0 / r dr r j(r) r 0 Now use the fact that j(r) = j0 a/r and integrate. For r < R we have B(r) = µ 0 r j 0ar = µ 0 j 0 a independent of radius r. To create a magnetic field of 2T inside of a solenoid of length 50cm and radius 1cm, how many total windings are required for the entire length if the current in the wire cannot exceed 10A? (1) 80,000 (2) 160,000 (3) 1,600 (4) 10,000 (5) 800,000 Problem 9 The field inside of a solenoid can be found by Ampere s Law and is: B = µ 0 ni = µ 0 i N l ( ) N = Bl µ 0 i = π 10 7 ( ) = 80,000 A proton of velocity m/s and mass kg is to be kept in a circular orbit using a magnet. What is the minimum diameter of the magnet necessary assuming that it provides a uniform magnetic field of 0.5T between its pole tips. (1) 2.5m (2) 1.2m (3) m (4) 1.6m (5) 0.8m The relation between the momentum and the radius of curvature of a charged particle in a uniform magnetic field is given by: mv = qbr d = 2r = 2mv qb d = 2.5m ( )( ) ( )( 0.5) = r 0

6 Problem 10 An initially uncharged capacitor and a 100 kω resistor are connected in series with an EMF source at t=0 to form a complete circuit. If the capacitor is charged to half of its maximum value in 0.1s, what must be the capacitance of the capacitor? (1) 1.4 μf (2) 1.0 μf (3) F (4) 0.5 μf (5) 0.5F The charge on the capacitor in an RC circuit increases in time as: q t ( ) where τ RC = RC ( ) = Cε 1 e t/τ RC this reaches half of its maximum when ( ) q( t) = 1 Cε = Cε 1 e t/rc 2 Problem 11 e t/rc = 1 2 t RC = ln2 t C = Rln2 = ln2 = 1.4µF Consider a circular loop of current as shown in the figure, where the distance r to point P along the axis of the loop is much larger than the loop radius R. If the distance r to point P is doubled, what is the ratio of the magnitude of the magnetic field at the new location of P to that at the original position? (1) 1/8 (2) 8 (3) ¼ (4) 4 (5) 1/2 For r R the magnetic field of a current loop becomes that of a magnetic dipole with the following dependence: B µ 0µ r 3 where μ is the magnetic dipole moment ia. Since this field falls off as r - 3, when the distance is doubled, the field magnitude is 1/8 of the initial value.

7 Problem 12 In the figure shown, a metal rod is pulled at a constant velocity along two parallel metal rails, connected with a strip of metal at one end. A magnetic field of magnitude B = 0.50T points out of the page. If the metal rod has a length L = 0.25m and a resistance of 0.3 Ω, what is the force necessary (in N) to pull the rod with a velocity of 2m/s? The rails and connector have negligible resistance. (1) 0.1 (2) 0.25 (3) 0.8 (4) 0.03 (5) 2.0 The magnitude of the induced EMF in the loop is ε = vlb from Faraday s law, since the amount of magnetic flux contained by the circuit is increasing. Since the flux is increasing, Lenz s Law says that the direction of the induced EMF and current will be to oppose that decrease, so the induced magnetic field should point into the page also. This requires a clockwise direction. The magnitude of the current in the rod is i = vlb R With a clockwise current, the current travels up the rod. This leads to a force pointing to the right: F = il B, with magnitude: F = vl2 B 2 R = 2( 0.25 ( )2 0.5) 2 = 0.10 N 0.3 Problem 13 The magnetic field passing perpendicularly through a square conducting loop of side length 2m as shown decreases linearly from 2T to 0 in 100 seconds. What is the induced EMF and the direction of the induced current?

8 (1) 0.08V, counter- clockwise (2) 0.08V, clockwise (3) 0.16V, counter- clockwise (4) 0.16V, clockwise (5) 0.02V, clockwise Since the magnetic flux is decreasing, Lenz s Law says the induced field will oppose that change. That leads to a counter- clockwise current in order to induce a field coming out of the page. By Faraday s Law: ε = d dt ΔB ( BA) = s 2 Δt = = 0.08V Problem 14 The current through an inductor with inductance L=0.3H is shown by the graph, with the direction from left to right through the inductor as shown. What is the EMF across the inductor (VL= Vright - Vleft), including sign, at t = 1ms? (1) 900 V (2) V (3) 1500 V (4) V (5) - 2 V Problem 15 The induced EMF on an inductor is given by: ε = L di dt At t=1ms, di dt = ( 2 8) = 3000 A s ( ) = 6 So the EMF is V L = ( ) = 900 V The switch S in the shown RL circuit is closed at time t=0. If L=10H, R=2 Ω, and ε = 6V, what is the magnitude of the potential difference across the inductor immediately after closing the switch.

9 (1) 6 V (2) 0 V (3) 3 V (4) 5 V (5) 30 V The current increases in an RL circuit as: ( ) where τ L = L R i( t) = ε R 1 e t/t L when t=0 then i=0, so there is no potential difference across the resistor. Thus by the loop rule the potential difference across the inductor must be equal to the EMF of 6V. Problem 16 Two wires are made of the same material. The second wire has twice the radius and twice the length of the first wire. What is the ratio of the second wire's resistance to the first one? (1) ½ (2) 2 (3) 8 (4) 1/8 (5) 1 If the first wire has length L and radius r, its resistance is R 1 = ρl / (πr 2 ), where ρ is the resistivity. Hence the resistance of the second wire is R 2 = ρ2l / (4πr 2 ) = R 1 / 2 Problem 17 A 5 Ohm resistor dissipates 10 Watts of power. What is the current i (in amps) through the resistor and the voltage V (in volts) across it? (1) i = 2, V = 5 2 (2) i = 5 2, V = 2 (3) i = 2, V = 5 (4) i = 5, V = 2 (5) i = 10, V = 10)

10 Problem 18 The power dissipated by a current i flowing through a resistance R is P = R. Hence the current is i = P / R = 10 W / 5Ω = 2 A. The voltage across the same resistor is V = ir = 5 2 V In the figure, R = 14.0Ω, C = 6.20 μf, and L = 54.0 mh, and the ideal battery has emf ε =34.0 V. The switch is kept at a for a long time and then thrown to position b. What are the frequency f (not angular frequency!) in Hz and the amplitude of the current oscillation I in Amps? (1) f = 275, i =.364 (2) f = 1730, i =.0580 (3) f = 43.8, i = 2.29 (4) f = 275, i = 2.43 (5) f = 1730, i = 2.43 This was problem from the assigned homework! Keeping the switch at a for a long time fully charges the capacitor to charge q 0 = CE = µc. When we then flip the switch to b at time t = 0 then initial current is i0 = 0 and the charge at any time is given by q(t) = q 0 cos(ωt), where the angular frequency is ω = 1/ LC. The frequency is therefore f = ω / 2π ~ 275 Hz. The current is i(t) = dq / dt = ωq 0 sin(ωt), hence the amplitude of the current oscillation is ωq 0 ~.364 A. Problem 19 An LC circuit with capacitance C = 4 çf and inductance L = 2.5 mh carries a total energy of U = 10 mj. At a particular instant the current through the inductor is i = 2 A. What is the charge on the capacitor at the same instant? (1) 200 μc (2) 283 μc (3) 400 μc (4) 141 μc (5) 566 μc Recall that the energy stored in a capacitor is U E = q 2 / 2C, and the energy stored in an inductor is U B = 1 2 Li2. Hence the charge on the capacitor is q = 2CU CL = C

11 Problem 20 What resistance R should be connected in series with an inductance L = 220 mh and capacitance C = 12.0 μ F for the maximum charge on the capacitor to decay to 99.0% of its initial value in 50.0 cycles? (Assume ω' ω.) (1) 8.66 mω (2) 4.33 mω (3) 54.5 mω (4) 3.76 mω (5) 1.88 mω This was problem from the assigned homework! Recall that the time evolutions of the capacitor charge in an RLC circuit takes the form q(t) = Qe Rt/2 L cos( ω t + φ). The time for fifty cycles is t = 50 2π / ω 100π LC. Having the charge decay to 99% of its maximum value after 50 cycles means e Rt/2 L =.99, which implies R L / C ln(100 / 99) / 50π ~ Ω.

### Solutions to PHY2049 Exam 2 (Nov. 3, 2017)

Solutions to PHY2049 Exam 2 (Nov. 3, 207) Problem : In figure a, both batteries have emf E =.2 V and the external resistance R is a variable resistor. Figure b gives the electric potentials V between the

### PHYS 241 EXAM #2 November 9, 2006

1. ( 5 points) A resistance R and a 3.9 H inductance are in series across a 60 Hz AC voltage. The voltage across the resistor is 23 V and the voltage across the inductor is 35 V. Assume that all voltages

### Louisiana State University Physics 2102, Exam 3 April 2nd, 2009.

PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 3 April 2nd, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),

### Exam 2 Solutions. ε 3. ε 1. Problem 1

Exam 2 Solutions Problem 1 In the circuit shown, R1=100 Ω, R2=25 Ω, and the ideal batteries have EMFs of ε1 = 6.0 V, ε2 = 3.0 V, and ε3 = 1.5 V. What is the magnitude of the current flowing through resistor

### Handout 10: Inductance. Self-Inductance and inductors

1 Handout 10: Inductance Self-Inductance and inductors In Fig. 1, electric current is present in an isolate circuit, setting up magnetic field that causes a magnetic flux through the circuit itself. This

### PHY 131 Review Session Fall 2015 PART 1:

PHY 131 Review Session Fall 2015 PART 1: 1. Consider the electric field from a point charge. As you move farther away from the point charge, the electric field decreases at a rate of 1/r 2 with r being

### Final on December Physics 106 R. Schad. 3e 4e 5c 6d 7c 8d 9b 10e 11d 12e 13d 14d 15b 16d 17b 18b 19c 20a

Final on December11. 2007 - Physics 106 R. Schad YOUR NAME STUDENT NUMBER 3e 4e 5c 6d 7c 8d 9b 10e 11d 12e 13d 14d 15b 16d 17b 18b 19c 20a 1. 2. 3. 4. This is to identify the exam version you have IMPORTANT

### Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits

Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the timevarying

### Exam 2 Solutions. Applying the junction rule: i 1 Applying the loop rule to the left loop (LL), right loop (RL), and the full loop (FL) gives:

PHY61 Eam Solutions 1. [8 points] In the circuit shown, the resistance R 1 = 1Ω. The batter voltages are identical: ε1 = ε = ε3 = 1 V. What is the current (in amps) flowing through the middle branch from

### Physics 102 Spring 2006: Final Exam Multiple-Choice Questions

Last Name: First Name: Physics 102 Spring 2006: Final Exam Multiple-Choice Questions For questions 1 and 2, refer to the graph below, depicting the potential on the x-axis as a function of x V x 60 40

### Exam 2, Phy 2049, Spring Solutions:

Exam 2, Phy 2049, Spring 2017. Solutions: 1. A battery, which has an emf of EMF = 10V and an internal resistance of R 0 = 50Ω, is connected to three resistors, as shown in the figure. The resistors have

### Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 1

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering.

### MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Spring 2014 Final Exam Equation Sheet. B( r) = µ o 4π

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2014 Final Exam Equation Sheet Force Law: F q = q( E ext + v q B ext ) Poynting Vector: S = ( E B) / µ 0 Force on Current Carrying

### Exam 3 November 19, 2012 Instructor: Timothy Martin

PHY 232 Exam 3 October 15, 2012 Exam 3 November 19, 2012 Instructor: Timothy Martin Student Information Name and section: UK Student ID: Seat #: Instructions Answer the questions in the space provided.

### Magnets. Domain = small magnetized region of a magnetic material. all the atoms are grouped together and aligned

Magnetic Fields Magnets Domain = small magnetized region of a magnetic material all the atoms are grouped together and aligned Magnets Ferromagnetic materials domains can be forced to line up by applying

### Physics Jonathan Dowling. Final Exam Review

Physics 2102 Jonathan Dowling Physics 2102 Final Exam Review A few concepts: electric force, field and potential Electric force: What is the force on a charge produced by other charges? What is the force

### 21 MAGNETIC FORCES AND MAGNETIC FIELDS

CHAPTER 1 MAGNETIC FORCES AND MAGNETIC FIELDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1 (d) Right-Hand Rule No 1 gives the direction of the magnetic force as x for both drawings A and B In drawing C, the

### MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Spring 2013 Exam 3 Equation Sheet. closed fixed path. ! = I ind.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 013 Exam 3 Equation Sheet Force Law: F q = q( E ext + v q B ext ) Force on Current Carrying Wire: F = Id s " B # wire ext Magnetic

### PHY2049 Fall11. Final Exam Solutions (1) 700 N (2) 350 N (3) 810 N (4) 405 N (5) 0 N

Exam Solutions 1. Three charges form an equilateral triangle of side length d = 2 cm. The top charge is q3 = 3 μc, while the bottom two are q1 = q2 = - 6 μc. What is the magnitude of the net force acting

### Yell if you have any questions

Class 36: Outline Hour 1: Concept Review / Overview PRS Questions Possible Exam Questions Hour : Sample Exam Yell if you have any questions P36-1 Before Starting All of your grades should now be posted

### Where k = 1. The electric field produced by a point charge is given by

Ch 21 review: 1. Electric charge: Electric charge is a property of a matter. There are two kinds of charges, positive and negative. Charges of the same sign repel each other. Charges of opposite sign attract.

### Louisiana State University Physics 2102, Exam 2, March 5th, 2009.

PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 2, March 5th, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),

### Yell if you have any questions

Class 31: Outline Hour 1: Concept Review / Overview PRS Questions possible exam questions Hour : Sample Exam Yell if you have any questions P31 1 Exam 3 Topics Faraday s Law Self Inductance Energy Stored

### Exam 3 Topics. Displacement Current Poynting Vector. Faraday s Law Self Inductance. Circuits. Energy Stored in Inductor/Magnetic Field

Exam 3 Topics Faraday s Law Self Inductance Energy Stored in Inductor/Magnetic Field Circuits LR Circuits Undriven (R)LC Circuits Driven RLC Circuits Displacement Current Poynting Vector NO: B Materials,

### Physics 2020 Exam 2 Constants and Formulae

Physics 2020 Exam 2 Constants and Formulae Useful Constants k e = 8.99 10 9 N m 2 /C 2 c = 3.00 10 8 m/s ɛ = 8.85 10 12 C 2 /(N m 2 ) µ = 4π 10 7 T m/a e = 1.602 10 19 C h = 6.626 10 34 J s m p = 1.67

### Figure 1 A) 2.3 V B) +2.3 V C) +3.6 V D) 1.1 V E) +1.1 V Q2. The current in the 12- Ω resistor shown in the circuit of Figure 2 is:

Term: 13 Wednesday, May 1, 014 Page: 1 Q1. What is the potential difference V B -V A in the circuit shown in Figure 1 if R 1 =70.0 Ω, R=105 Ω, R 3 =140 Ω, ε 1 =.0 V and ε =7.0 V? Figure 1 A).3 V B) +.3

### EXAM 3: SOLUTIONS. B = B. A 2 = BA 2 cos 0 o = BA 2. =Φ(2) B A 2 = A 1 cos 60 o = A 1 2 =0.5m2

EXAM : S Q.The normal to a certain m area makes an angle of 6 o with a uniform magnetic field. The magnetic flux through this area is the same as the flux through a second area that is perpendicular to

### The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:

PHYS 102 Exams Exam 2 PRINT (A) The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d: It is connected to a battery with constant emf V.

### Physics 208, Spring 2016 Exam #3

Physics 208, Spring 206 Exam #3 A Name (Last, First): ID #: Section #: You have 75 minutes to complete the exam. Formulae are provided on an attached sheet. You may NOT use any other formula sheet. You

### a. Clockwise. b. Counterclockwise. c. Out of the board. d. Into the board. e. There will be no current induced in the wire

Physics 1B Winter 2012: Final Exam For Practice Version A 1 Closed book. No work needs to be shown for multiple-choice questions. The first 10 questions are the makeup Quiz. The remaining questions are

### Physics 2B Spring 2010: Final Version A 1 COMMENTS AND REMINDERS:

Physics 2B Spring 2010: Final Version A 1 COMMENTS AND REMINDERS: Closed book. No work needs to be shown for multiple-choice questions. 1. A charge of +4.0 C is placed at the origin. A charge of 3.0 C

### Last time. Ampere's Law Faraday s law

Last time Ampere's Law Faraday s law 1 Faraday s Law of Induction (More Quantitative) The magnitude of the induced EMF in conducting loop is equal to the rate at which the magnetic flux through the surface

### Quiz 4 (Discussion Session) Phys 1302W.400 Spring 2018

Quiz 4 (Discussion ession) Phys 1302W.400 pring 2018 This group quiz consists of one problem that, together with the individual problems on Friday, will determine your grade for quiz 4. For the group problem,

### Physics 2135 Exam 2 October 20, 2015

Exam Total / 200 Physics 2135 Exam 2 October 20, 2015 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. 1. A straight wire segment

### Inductance, RL and RLC Circuits

Inductance, RL and RLC Circuits Inductance Temporarily storage of energy by the magnetic field When the switch is closed, the current does not immediately reach its maximum value. Faraday s law of electromagnetic

### Physics 1302W.400 Lecture 33 Introductory Physics for Scientists and Engineering II

Physics 1302W.400 Lecture 33 Introductory Physics for Scientists and Engineering II In today s lecture, we will discuss generators and motors. Slide 30-1 Announcement Quiz 4 will be next week. The Final

This test covers Faraday s Law of induction, motional emf, Lenz s law, induced emf and electric fields, eddy currents, self-inductance, inductance, RL circuits, and energy in a magnetic field, with some

### Chapter 32 Solutions

Chapter 3 Solutions *3. ε = L I t = (3.00 0 3 H).50 A 0.00 A 0.00 s =.95 0 V = 9.5 mv 3. Treating the telephone cord as a solenoid, we have: L = µ 0 N A l = (4π 0 7 T m/a)(70.0) (π)(6.50 0 3 m) 0.600 m

### Exam 2 Solutions. Prof. Darin Acosta Prof. Greg Stewart March 27, b 5Ω ->i 1 <- i 2 5Ω. 3V 10Ω 6 V - - i 3 d c 10Ω

PHY49 Spring 6 Exam Solutions Prof. Darin Acosta Prof. Greg Stewart March 7, 6 Exam Solutions a b 5Ω ->i 1

### ELECTROMAGNETIC OSCILLATIONS AND ALTERNATING CURRENT

Chapter 31: ELECTROMAGNETIC OSCILLATIONS AND ALTERNATING CURRENT 1 A charged capacitor and an inductor are connected in series At time t = 0 the current is zero, but the capacitor is charged If T is the

### PHYS102 Previous Exam Problems. Induction

PHYS102 Previous Exam Problems CHAPTER 30 Induction Magnetic flux Induced emf (Faraday s law) Lenz law Motional emf 1. A circuit is pulled to the right at constant speed in a uniform magnetic field with

### PHYS General Physics for Engineering II FIRST MIDTERM

Çankaya University Department of Mathematics and Computer Sciences 2010-2011 Spring Semester PHYS 112 - General Physics for Engineering II FIRST MIDTERM 1) Two fixed particles of charges q 1 = 1.0µC and

### Phys102 Final-132 Zero Version Coordinator: A.A.Naqvi Wednesday, May 21, 2014 Page: 1

Coordinator: A.A.Naqvi Wednesday, May 1, 014 Page: 1 Q1. What is the potential difference V B -V A in the circuit shown in Figure 1 if R 1 =70.0 Ω, R =105 Ω, R 3 =140 Ω, ε 1 =.0 V and ε =7.0 V? A).3 V

### Principles of Physics II

Principles of Physics II J. M. Veal, Ph. D. version 18.05.4 Contents 1 Fluid Mechanics 3 1.1 Fluid pressure............................ 3 1. Buoyancy.............................. 3 1.3 Fluid flow..............................

### = e = e 3 = = 4.98%

PHYS 212 Exam 2 - Practice Test - Solutions 1E In order to use the equation for discharging, we should consider the amount of charge remaining after three time constants, which would have to be q(t)/q0.

### PRACTICE EXAM 1 for Midterm 2

PRACTICE EXAM 1 for Midterm 2 Multiple Choice Questions 1) The figure shows three identical lightbulbs connected to a battery having a constant voltage across its terminals. What happens to the brightness

### Part 4: Electromagnetism. 4.1: Induction. A. Faraday's Law. The magnetic flux through a loop of wire is

1 Part 4: Electromagnetism 4.1: Induction A. Faraday's Law The magnetic flux through a loop of wire is Φ = BA cos θ B A B = magnetic field penetrating loop [T] A = area of loop [m 2 ] = angle between field

### A) I B) II C) III D) IV E) V

1. A square loop of wire moves with a constant speed v from a field-free region into a region of uniform B field, as shown. Which of the five graphs correctly shows the induced current i in the loop as

### Physics 2220 Fall 2010 George Williams THIRD MIDTERM - REVIEW PROBLEMS

Physics 2220 Fall 2010 George Williams THIRD MIDTERM - REVIEW PROBLEMS Solution sets are available on the course web site. A data sheet is provided. Problems marked by "*" do not have solutions. 1. An

### Name (Last, First): You may use only scientific or graphing calculators. In particular you may not use the calculator app on your phone or tablet!

Final Exam : Physics 2113 Fall 2014 5:30PM MON 8 DEC 2014 Name (Last, First): Section # Instructor s name: Answer all 6 problems & all 8 questions. Be sure to write your name. Please read the questions

### Exam 3 Solutions. The induced EMF (magnitude) is given by Faraday s Law d dt dt The current is given by

PHY049 Spring 008 Prof. Darin Acosta Prof. Selman Hershfield April 9, 008. A metal rod is forced to move with constant velocity of 60 cm/s [or 90 cm/s] along two parallel metal rails, which are connected

### University of the Philippines College of Science PHYSICS 72. Summer Second Long Problem Set

University of the Philippines College of Science PHYSICS 72 Summer 2012-2013 Second Long Problem Set INSTRUCTIONS: Choose the best answer and shade the corresponding circle on your answer sheet. To change

### 1 2 U CV. K dq I dt J nqv d J V IR P VI

o 5 o T C T F 3 9 T K T o C 73.5 L L T V VT Q mct nct Q F V ml F V dq A H k TH TC L pv nrt 3 Ktr nrt 3 CV R ideal monatomic gas 5 CV R ideal diatomic gas w/o vibration V W pdv V U Q W W Q e Q Q e Carnot

### Chapter 32. Inductance

Chapter 32 Inductance Joseph Henry 1797 1878 American physicist First director of the Smithsonian Improved design of electromagnet Constructed one of the first motors Discovered self-inductance Unit of

### Chapter 30 Inductance

Chapter 30 Inductance In this chapter we investigate the properties of an inductor in a circuit. There are two kinds of inductance mutual inductance and self-inductance. An inductor is formed by taken

### Final Exam: Physics Spring, 2017 May 8, 2017 Version 01

Final Exam: Physics2331 - Spring, 2017 May 8, 2017 Version 01 NAME (Please Print) Your exam should have 11 pages. This exam consists of 18 multiple-choice questions (2 points each, worth 36 points), and

### Questions A hair dryer is rated as 1200 W, 120 V. Its effective internal resistance is (A) 0.1 Ω (B) 10 Ω (C) 12Ω (D) 120 Ω (E) 1440 Ω

Questions 4-41 36. Three 1/ µf capacitors are connected in series as shown in the diagram above. The capacitance of the combination is (A).1 µf (B) 1 µf (C) /3 µf (D) ½ µf (E) 1/6 µf 37. A hair dryer is

### 18 - ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS ( Answers at the end of all questions ) Page 1

( Answers at the end of all questions ) Page ) The self inductance of the motor of an electric fan is 0 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of 8 µ F (

### Physics GRE: Electromagnetism. G. J. Loges 1. University of Rochester Dept. of Physics & Astronomy. xkcd.com/567/

Physics GRE: Electromagnetism G. J. Loges University of Rochester Dept. of Physics & stronomy xkcd.com/567/ c Gregory Loges, 206 Contents Electrostatics 2 Magnetostatics 2 3 Method of Images 3 4 Lorentz

### Name (Print): 4 Digit ID: Section:

Physics 11 Sample Common Exam 3: Sample 5 Name (Print): 4 Digit ID: Section: Honors Code Pledge: As an NJIT student I, pledge to comply with the provisions of the NJIT Academic Honor Code. I assert that

### Physics 420 Fall 2004 Quiz 1 Wednesday This quiz is worth 6 points. Be sure to show your work and label your final answers.

Quiz 1 Wednesday This quiz is worth 6 points. Be sure to show your work and label your final answers. 1. A charge q 1 = +5.0 nc is located on the y-axis, 15 µm above the origin, while another charge q

### Magnetic Fields; Sources of Magnetic Field

This test covers magnetic fields, magnetic forces on charged particles and current-carrying wires, the Hall effect, the Biot-Savart Law, Ampère s Law, and the magnetic fields of current-carrying loops

### Exam 2 Solutions. PHY2054 Spring Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014

Exam 2 Solutions Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014 1. A series circuit consists of an open switch, a 6.0 Ω resistor, an uncharged 4.0 µf capacitor and a battery with emf 15.0 V and internal

### force per unit length

Physics 153 Sample Examination for Fourth Unit As you should know, this unit covers magnetic fields, how those fields interact with charged particles, how they are produced, how they can produce electric

### cancel each other out. Thus, we only need to consider magnetic field produced by wire carrying current 2.

PC1143 2011/2012 Exam Solutions Question 1 a) Assumption: shells are conductors. Notes: the system given is a capacitor. Make use of spherical symmetry. Energy density, =. in this case means electric field

### Yell if you have any questions

Class 36: Outline Hour 1: Concept Review / Overview PRS Questions Possible Exam Questions Hour : Sample Exam Yell if you have any questions P36-1 efore Starting All of your grades should now be posted

### ELECTRO MAGNETIC INDUCTION

ELECTRO MAGNETIC INDUCTION 1) A Circular coil is placed near a current carrying conductor. The induced current is anti clock wise when the coil is, 1. Stationary 2. Moved away from the conductor 3. Moved

### Exam II. Solutions. Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points. The wires repel each other.

Exam II Solutions Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points. 1.! Concerning electric and magnetic fields, which of the following is wrong?!! A

### Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current.

Inductance Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current. Basis of the electrical circuit element called an

### PHY 2049 SPRING 2001 EXAM 3

PHY 2049 SPING 2001 EXAM 3 1. A negatively charged particle moves with velocity v in a uniform magnetic field as shown in figure (both vector and vector v are in the plane of the paper). What is the direction

### PHYS 212 Final Exam (Old Material) Solutions - Practice Test

PHYS 212 Final Exam (Old Material) Solutions - Practice Test 1E If the ball is attracted to the rod, it must be made of a conductive material, otherwise it would not have been influenced by the nearby

### Physics 106, Section 1

Physics 106, Section 1 Magleby Exam 2, Summer 2012 Exam Cid You are allowed a pencil and a testing center calculator. No scratch paper is allowed. Testing center calculators only. 1. A circular coil lays

### Circuits Capacitance of a parallel-plate capacitor : C = κ ε o A / d. (ρ = resistivity, L = length, A = cross-sectional area) Resistance : R = ρ L / A

k = 9.0 x 109 N m2 / C2 e = 1.60 x 10-19 C ε o = 8.85 x 10-12 C2 / N m2 Coulomb s law: F = k q Q / r2 (unlike charges attract, like charges repel) Electric field from a point charge : E = k q / r2 ( towards

### SUMMARY Phys 2523 (University Physics II) Compiled by Prof. Erickson. F e (r )=q E(r ) dq r 2 ˆr = k e E = V. V (r )=k e r = k q i. r i r.

SUMMARY Phys 53 (University Physics II) Compiled by Prof. Erickson q 1 q Coulomb s Law: F 1 = k e r ˆr where k e = 1 4π =8.9875 10 9 N m /C, and =8.85 10 1 C /(N m )isthepermittivity of free space. Generally,

### Physics 6B Summer 2007 Final

Physics 6B Summer 2007 Final Question 1 An electron passes through two rectangular regions that contain uniform magnetic fields, B 1 and B 2. The field B 1 is stronger than the field B 2. Each field fills

### Exam 2 Solutions. Answer: 3.0 W Solution: The total current is in the series circuit is 1 A, so the power dissipated in R 2 is i 2 R 2

Exam 2 Solutions Prof. Pradeep Kumar Prof. Paul Avery Mar. 21, 2012 1. A portable CD player does not have a power rating listed, but it has a label stating that it draws a maximum current of 159.0 ma.

### Version 001 CIRCUITS holland (1290) 1

Version CIRCUITS holland (9) This print-out should have questions Multiple-choice questions may continue on the next column or page find all choices before answering AP M 99 MC points The power dissipated

### Chapter 31. Faraday s Law

Chapter 31 Faraday s Law 1 Ampere s law Magnetic field is produced by time variation of electric field B s II I d d μ o d μo με o o E ds E B Induction A loop of wire is connected to a sensitive ammeter

### Physics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe. Useful Information. Your name sticker. with exam code

Your name sticker with exam code Physics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe SIGNATURE: 1. The exam will last from 4:00 p.m. to 7:00 p.m. Use a #2 pencil to make entries on the

### Chapter In Fig , the magnetic flux through the loop increases according to the relation Φ B. =12.0t

Chapter 30 30.1 In Fig. 30-37, the magnetic lux through the loop increases according to the relation = 6.0t 2 + 7.0t where the lux is in milliwebers and t is in seconds. (a) What is the magnitude o the

### Electromagnetic Induction (Chapters 31-32)

Electromagnetic Induction (Chapters 31-3) The laws of emf induction: Faraday s and Lenz s laws Inductance Mutual inductance M Self inductance L. Inductors Magnetic field energy Simple inductive circuits

### ELECTROMAGNETIC INDUCTION AND FARADAY S LAW

ELECTROMAGNETIC INDUCTION AND FARADAY S LAW Magnetic Flux The emf is actually induced by a change in the quantity called the magnetic flux rather than simply py by a change in the magnetic field Magnetic

### = 8.89x10 9 N m 2 /C 2

PHY303L Useful Formulae for Test 2 Magnetic Force on a moving charged particle F B = q v B Magnetic Force on a current carrying wire F B = i L B Magnetic dipole moment µ = NiA Torque on a magnetic dipole:

### Chapter 28 Solutions

Chapter 8 Solutions 8.1 (a) P ( V) R becomes 0.0 W (11.6 V) R so R 6.73 Ω (b) V IR so 11.6 V I (6.73 Ω) and I 1.7 A ε IR + Ir so 15.0 V 11.6 V + (1.7 A)r r 1.97 Ω Figure for Goal Solution Goal Solution

### Physics 227 Final Exam Wednesday, May 9, Code: 000

Physics 227 Final Exam Wednesday, May 9, 2018 Physics 227, Section RUID: Code: 000 Your name with exam code Your signature Turn off and put away LL electronic devices NOW. NO cell phones, NO smart watches,

### PS I AP Physics 2 Electromagnetic Induction Multiple Choice Questions

PS I AP Physics 2 Electromagnetic Induction Multiple Choice Questions 1. A beam of electrons travels between two parallel coils of wire, as shown in the figures above. When the coils do not carry a current,

### Chapter 20: Electromagnetic Induction. PHY2054: Chapter 20 1

Chapter 20: Electromagnetic Induction PHY2054: Chapter 20 1 Electromagnetic Induction Magnetic flux Induced emf Faraday s Law Lenz s Law Motional emf Magnetic energy Inductance RL circuits Generators and

### /20 /20 /20 /60. Dr. Galeazzi PHY207 Test #3 November 20, I.D. number:

Signature: Name: I.D. number: You must do ALL the problems Each problem is worth 0 points for a total of 60 points. TO GET CREDIT IN PROBLEMS AND 3 YOU MUST SHOW GOOD WORK. CHECK DISCUSSION SECTION ATTENDED:

### Physics 1c Practical, Spring 2015 Hw 3 Solutions

Physics 1c Practical, Spring 2015 Hw 3 Solutions April 16, 2015 1 Serway 31.79 (5 points) By Lenz s Law, current will flow in each of the two sub-loops to oppose the change in flux. One can easily see

### Induction_P1. 1. [1 mark]

Induction_P1 1. [1 mark] Two identical circular coils are placed one below the other so that their planes are both horizontal. The top coil is connected to a cell and a switch. The switch is closed and

### Physics 212 Midterm 2 Form A

1. A wire contains a steady current of 2 A. The charge that passes a cross section in 2 s is: A. 3.2 10-19 C B. 6.4 10-19 C C. 1 C D. 2 C E. 4 C 2. In a Physics 212 lab, Jane measures the current versus

### Physics Physics 2102

Physics 2102 Jonathan Dowling Physics 2102 Exam 2: Review Session Chapters 24.9-28.8 / HW04-06 Some links on exam stress: http://appl003.lsu.edu/slas/cas.nsf/\$content/stress+management+tip+1 http://wso.williams.edu/orgs/peerh/stress/exams.html

### Physics 2135 Exam 2 March 22, 2016

Exam Total Physics 2135 Exam 2 March 22, 2016 Key Printed Name: 200 / 200 N/A Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. B 1. An air-filled

### PHYS 1444 Section 02 Review #2

PHYS 1444 Section 02 Review #2 November 9, 2011 Ian Howley 1 1444 Test 2 Eq. Sheet Terminal voltage Resistors in series Resistors in parallel Magnetic field from long straight wire Ampére s Law Force on

### Physics 2212 GH Quiz #4 Solutions Spring 2016

Physics 2212 GH Quiz #4 Solutions Spring 2016 I. (18 points) A bar (mass m, length L) is connected to two frictionless vertical conducting rails with loops of wire, in the presence of a uniform magnetic

### Electromagnetic Induction Faraday s Law Lenz s Law Self-Inductance RL Circuits Energy in a Magnetic Field Mutual Inductance

Lesson 7 Electromagnetic Induction Faraday s Law Lenz s Law Self-Inductance RL Circuits Energy in a Magnetic Field Mutual Inductance Oscillations in an LC Circuit The RLC Circuit Alternating Current Electromagnetic