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1 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 I have not violated the NJIT Academic Honor Code. Instructions: e sure your name and section number are on both the Scantron form and the question book. There are 18 multiple choice questions on this exam, all worth the same amount. Sixteen correct answers yield a score of 100%. Extra credit questions 17, and 18 may be harder than the others. Answer each question on the Scantron sheet using a # pencil. Do your calculations on the exam papers under each question: use the backs of pages for extra space. A default formula sheet is provided at the end of this exam booklet. It is the same as the one made available on the course web site. Otherwise, this is a closed book exam. Make sure to bring your own calculator: you will need it and sharing of calculators is not permitted. If you have questions or need something call your proctor or instructor. As you know, NJIT has a zerotolerance policy for ethics code violations. Students are not to communicate with each other once the exam has started. All cell phones, pagers, or similar electronic devices must be turned off. 1. In the multiloop circuit shown in the figure, both batteries are ideal EMFs with V 1 =.0 Volt, and V = 3.0 Volts. The resistances R 1 = 5.0 Ω, and R = 4.0 Ω. Find the current I through R. A) 1.0 A, down ) 1.0 A, up C) 0.5 A, up D).0 A, down E) 0.5 A, down V I R I R 1 I For the circuit of the previous problem find the current I 1 through the battery labeled V 1 A) 0.3 A, down ) 0.3 A, up C) 0. A, down D) 0. A, up E) 0.4 A, up Page 1
2 3. The batteries in the sketch have Emfs of V 1 = 40 Volts and V = 60 Volts. They have zero internal resistance. The resistances are R 1 = 15 Ω and R = 10 Ω. Find the magnitude and direction of the current in R 1. A) 1.0 A clockwise ) 0.8 A counterclockwise C) 4.0 A clockwise D) 4.0 A counterclockwise E) 0.8 A clockwise  + R V +  R 1 4. Find the power expended by the battery on the left in the previous circuit (a negative sign implies that power is actually consumed by the battery as for example when it is being charged). A. 160 W. 30 W C. 0 W D. 40 W E. 80 W 5. A fast positively charged particle, which moves from right to left in the plane of the page, enters a region of magnetic field which points directly into the page. The particle will: A) be deflected out of the plane of the page ) continue moving in a straight line, but will accelerate C) continue moving in a straight line, but will decelerate D) be deflected down, still in the plane of the page E) be deflected up, still in the plane of the page 6. An electron is moving with a velocity of v = 10 6 m/s in the positive x direction. It is crossing a region in which a uniform magnetic field with = 50,000 Teslas points in the negative y direction. Find the force acting on the particle. A) 4.0 x 109 N, +z direction ) 8.0 x 109 N, z direction C) 8.0 x 109 N, +z direction D) 8.0 x 109 N, y direction E) 8.0 x 109 N, +y direction Page
3 7. The sketch at right shows a series RC circuit. The applied EMF E = 4 V., R = 0 MΩ, and C = 5 µf. The capacitor is initially uncharged. After the switch is closed at point a, how long does it take for the charging current through the resistor to fall to 0% of its initial value? Select the closest answer: A) 160 sec ) 30 sec C) 1.6 sec D) 3. sec E) 60 sec ε a b C R 8. Using the same circuit as the previous problem. Suppose the switch is thrown to b after a long time when the capacitor is fully charged. How long will it take for the charge on the capacitor to drop to 0% of its value when fully charged? A) 160 sec ) 30 sec C) 1.6 sec D) 3. sec E) 60 sec 9. The magnetic field is the same as in problem 6, but the velocity vector of the electron is in the xy plane with components (10 6, 10 6, 0) m/s. Find the force acting on the particle: A) none of the other choices ) 11.3 x 109 N, +z direction C) 6.0 x 109 N, x direction D) 5.7 x 109 N,  z direction E) 8.0 x 109 N, +z direction 10. A long straight wire lying on the xaxis carries a current of 10 A. in the positive xdirection. An electron located 10 cm from the origin on the positive yaxis is traveling parallel to the wire with a speed of x 10 5 m/s, also in the positive x direction. Assume the x and y axes point to the right and up respectively in the plane of the paper, with the z axis pointing out of the page toward the viewer. What force does the magnetic field of the current exert on the electron? A) 3. x N. along +z direction ) 3. x N. along y direction C) 6.4 x N. along +y direction D) 6.4 x N. along +z direction E) zero Page 3
4 11. A wire whose length L = 8 m is placed in a uniform magnetic field of magnitude = 0.5 T. which points out of the page. The wire carries a current of 5 A to the right in the plane of the page. What are the magnitude and direction of the force exerted on the wire by the magnetic field? A) 40 N, out of the page ) 0 N, up in the plane of the page C) 0 N, down in the plane of the page D) 40 N, up in the plane of the page E) N, down in the plane of the page 1. A proton is moving with a speed of 3x10 5 m/s in a uniform magnetic field of = 0.01 Tesla. The field points directly out of the page, and the velocity of the proton lies within in the plane of the page. Find the radius of the path the proton takes and the direction (CW or CCW). The mass of the proton is approximately 1.7x107 kg. The answer is closest to: A) 1.06 m, CCW ) 0.03m, CCW C) 0.3 m, CW D) 1.06 m, CW E) the proton moves in a straight line 13. The magnetic field lines near an infinitely long wire carrying a current: A) point away from the wire ) point toward the wire C) run parallel to the wire D) form a helix around the wire E) are concentric circles around the wire Page 4
5 14. Two wires are each 4 meters long and are parallel to the xaxis. They are separated by a perpendicular distance of 10 cm and carry currents of 50 A. and 100 A. in the positive x direction. Find the interaction force between the wires: A) zero ) 0.04 N, attraction C) 0.04 N, repulsion D) x 103 N, attraction E) x 103 N, repulsion 15. In the sketch (not drawn to scale) a long straight wire carries a downward pointing current I 1 = 7A in the plane of the page. A square frame whose sides a = 0.7 cm is also in the plane of the page and to the right of the straight wire. It carries a counterclockwise current of I = 6A and is oriented parallel to the wire. Which of the following correctly describes the net force on the frame: I 1 a a I A) the frame is attracted to the wire ) the frame is repelled by the wire C) the net force on the frame is zero, but there is a torque which attempts to turn it around a vertical axis D) the net force on the frame is zero, but there is a torque which attempts to turn it around a horizontal axis E) both the net force and torque are zero 16. The magnetic field at the center of a circular current loop is = 4.0 x 105 T, out of the page toward the viewer. The radius of the loop is 3.14 cm. The current flowing through the wire is closest to: A) 0.64 A. CCW ) A. CCW C). A. CW D) 0.64 A. CW E) Cannot be determined I Page 5
6 Extra Credit: Problems 17 and 18 use the sketch at the right, which shows a crosssection of a long coaxial cable. The current i 1 = 1 A in the center conductor, flowing out of the page (toward the viewer). The current i = 3 A. in the shield, flowing into the page (away from the viewer). The radius R of the shield is mm. Point a is 1 mm from the center. Point b is 3 mm from the center. center wire i Find the magnitude and direction of the magnetic field at point a. It may help to use Ampere s Law. The field is: A) 00 µt, counterclockwise ) 100 µt, clockwise C) 100 µt, counterclockwise D) 00 µt, clockwise E) Zero shield wire i a R b 18. Find the magnitude and direction of the magnetic field at point b. As before, it may help to use Ampere s Law. The field is: A) Zero ) 00 µt, clockwise C) 133 µt, counterclockwise D) 100 µt, clockwise E) 133 µt, clockwise Page 6
7 q = idt = i t Physics 11 Common Exam 3 Formulas Area of circle = πr Circumference of circle = πr 1 meter = 1000 mm = 100 cm 1 kg = 1000 g Surface area of sphere = 4πr Volume of sphere = (4/3)πr 3, 1 µc = 106 C 1 nc = 109 C 1/4πε o = k e = 9 x 10 9 Nm /C, ε o = 8.85 x 101 C /Nm, e = x C, m e = 9.11 x kg r 1 q q r 1 1 Q Point charges: F = rˆ E = rˆ where rˆ is a unit vector F = qe F 4 πε net = ma 0 r 4 πε 0 r r n r r r r Superposition: F = F + F + F E = force per unit charge at a point = 4 F net on 1 1,i 1, 1, 3 1, +... i= Shell Theorem (spheres only): mimics point charge outside, inside E or F is zero Dipole moment: p = qd τ dipole = pe U dipole = p.e dφ E = E. nda = EAcos(φ) Φ E = electric flux = q enc /ε o = E. da over a Gaussian surface σ = surface charge density E conducting sheet = σ/ε o E nonconducting sheet = σ/ε o E line = (1/πε o )λ/r V = U/q =  E cosθ ds =  E. ds U el = q V V = k e Q/r E =  dv/dx V V o = E(x x o ) Q = CV Electrostatic PE: U el = Q /C = CV / W = K + U C parallel = ΣC i 1/C series = Σ (1/C i ) C series = C 1 C /( C 1 +C ) C parallel plates = κε o A/d C sphere = 4πε o R Dielectric constant: C tot = κ C vac = dq = idt i = dq/dt i J d A = J A J = qnv drift J = σe σ = 1/ρ R = V / i V = ir R = ρl/a ρ = ρ 0 (1 + α(t T 0 )) Ohms Law: R independent of V R series = R 1 + R 1/R parallel = 1/R 1 + 1/ R R para = R 1 R /( R 1 + R ) P = du el /dt = iv P resistor = i R = V /R Junction rule: Σi in = Σi out Loop rule: Σ V i = 0 around any closed circuit path. Follow currents. RC Circuits: RC = time constant for circuit charging: i(t) = (V/R)e t/rc Q(t) = CV cap (t) = CV (1  e t/rc ) discharging: i(t) = (Q o /RC)e t/rc Q(t) = Q o e t/rc F m = q vx F e = qe F m = i Lx τ = µx U = µ. µ=nia normal to loop Cyclotron motion : r = mv/(q) period = πm/(q) ω = q/m f = ω /π = 1/period iot Savart: d = (µ o /4π)(i ds sinθ)/r where d is in the direction of ds x r µ o = 4π x 107 Tm/A wire = µ o i / πr arc = µ o iφ /4πR loop = µ o i /R solenoid =µ 0 in F = µ o i 1 i L /πd ( straight wires) Amperes law:.ds = µ 0 i enclosed for a closed Amperian loop on dipole axis Magnetic flux: dφ =.da Φ =.da Φ =Acos(θ) Φ = 0 over every Gaussian surface E p + πε 0 z 1 3 for large z Kinematics: v = v o + at x x o = v o t + ½at v = v o + a(x x o ) x x o = ½(v + v o )t Vectors: Components: a x = a cos(θ) a y = a sin(θ) a = a x i + a y j a = sqrt[ a x + a y ] θ = tan 1 (a y /a x ) Addition: a + b = c implies c x = a x + b x, c y = a y + b y Dot product: a b = a b cos(φ) = a x b x + a y b y + a z b z unit vectors: i i = j j = k k = 1; i j = i k = j k = 0 Cross product: a x b = a b sin(φ); c = a x b = (a y b z a z b y ) i + (a z b x a x b z ) j + (a x b y a y b x ) k a x b = b x a, a x a = 0 always; c = a x b is perpendicular to ab plane; if a b then a x b = 0 i x i = j x j = k x k = 0, i x j = k j x k = i k x i = j Page 7
8 ANSWER KEY Physics 11 Common Exam 3, Sample 5 1. E 0.5 A., down. 0.3 A., up 3. D 4.0 A. counterclockwise 4. A 160 Watts 5. D be deflected down, still in the plane of the page 6. C 8.0 x 109 N, +z direction 7. A 160 sec 8. A 160 sec 9. E 8.0 x 109 N, +z direction 10. C 6.4 x N. along +y direction 11. C 0 N, down in the plane of the page 1. C 0.3 m, CW 13. E are concentric circles around the wire N, attraction 15. A the frame is attracted to the wire 16. A., CCW 17. A 00 µt, counterclockwise 18. E 133 µt, clockwise Page 8
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