Physics 212 / Summer 2009 Name: ANSWER KEY Dr. Zimmerman Ch. 26 Quiz

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1 Physics 1 / Summer 9 Name: ANSWER KEY h. 6 Quiz As shown, there are three negatie charges located at the corners of a square of side. There is a single positie charge in the center of the square. (a) Draw arrows indicating the direction of the indiidual forces acting on the charge at the center. Be sure you gie the arrows the appropriate relatie size. (b) If the alue of Q= 5. n, and = 1 cm, determine the net force on the charge at the center and present your answer in component form. (c) What additional charge could be placed at the empty corner of the square so that the charge in the center is in equilibrium? Explain. Q Q + Q Q (b) By symmetry, two of the forces cancel. Thus we need only find the force due to the -Q charge, Q o ˆ o F = cos(45 ) i sin(45 ) ˆ 4 πε ( /) j Q ˆ ˆ 5 F = i + j = ( iˆ+ ˆj) N / πε (c) Placing another negatie charge (-Q) at the upper right hand corner would put the charge at the center in equilibrium since the forces would be equal and opposite and thus cancel.

2 Physics 1 / Summer 9 Name: ANSWER KEY h. 7 Quiz As shown, a thin rod of length is on the x-axis a distance from the origin and contains a total charge Q spread uniformly oer it. The goal is to find the electric field, E, at the origin ( x = ). (a) What is the linear charge density λ? (b) Determine the infinitesimal charge, dq. (c) Write down the expression for the infinitesimal electric field, de, at the origin. (d) Finally, calculate the integral with the correct limits of integration. dq x (a) λ = Q/ (b) Qdx dq = λdx = dq (c) de = ( iˆ ) x ( Q/ ) dx de = ( iˆ ) x (d) Q E = x= x= dx iˆ x Q 1 ˆ Q E = i = x 8πε i ˆ

3 Physics 1 / Summer 9 Name: ANSWER KEY h. 8 Quiz A conducting spherical shell of inner radius a and outer radius b surrounds a point charge + Q (which is fixed in place). Someone puts an additional charge + Q on the outer surface of the conductor (at r = b). (a) Draw and label the charge that must exist on each of the conductor surfaces at r = a and r = b in electrostatic equilibrium. (b) Using Gauss law, determine the electric field in three regions: r < a, a< r < b, and r > b. (c) What is the surface charge density ( η ) at r = a and r = b? a Q b We use spherical Gaussian surfaces concentric with the center of the sphere. (a) Q moes to the inner surface of the caity (at r = a), leaing behind +Q on the outer surface at r = b (b) r < a: The charge enclosed is Q, so E π r E 4 Q = = ε Q rˆ a< r < b: Here, the net charge enclosed is zero, Thus E = rˆ r > b: The charge enclosed is Q+ ( Q) + Q= Q, so (c) At r At r a, the surface charge density is, η Q =. 4π a Q 4π a = r= a = b, the surface charge density is, ηr= a= Q Q = = ε E π r E 4 rˆ

4 Physics 1 / Summer 9 Name: ANSWER KEY h. 9 Quiz A proton ( q p 19 = ), initially at rest, is released from a metal surface that is charged to a potential of 3 V. What is the speed of the proton when it is far from the surface? The mass of 7 a proton is m = kg. p Use conseration of energy: K + U = K + U i i f f + qv = m + 1 p i p f f qv V = = = 19 p i (1.6 1 )(3 ) 7 mp kg m / s

5 Physics 1 / Summer 9 Name: ANSWER KEY h. 3 Quiz What are (a) the charge on and (b) the potential difference across each capacitor in the following circuit? 1 =. μf, = 4. μf, 3 = 6. μf (a) and are in series so, F 1 3 = + μ = μ.4 5 F = μf 6.V 3 1 Then 1 and 3 are in parallel and will hae the same potential difference that of the battery. 3 Q = (6. V) = 1 μ 1 1 Q = (6. V) = 14.4 μ = Q = Q (Because and are in series they hae the same charge that of.) 3 3 (b) Δ V1 = 6.V ; it is in parallel with the battery Δ V = Q / = 3.6V Δ V = Q / =.4V (The oltage across and 3 should add to 6. V since they re in series.)

6 Physics 1 / Summer 9 Name: ANSWER KEY h. 31 Quiz The electron beam inside an analog teleision picture tube is.44 mm in diameter and carries a current of 5 μ A. The electron beam impinges on the inside of the picture tube screen. (a) How many electrons strike the screen each second? (b) What is the current density in the electron 7 beam? (c) The electrons moe with a elocity of 4. 1 m/ s. What electric field strength is needed to accelerate electrons from rest to this elocity in a distance of 5. mm? (d) Each electron transfers its energy to the picture tube screen upon impact. What is the power deliered to the screen by the electron beam? (a) Q Δt 5 5 I = Q= IΔ t = (5. 1 A)(1. s) = 5. 1 N electron 5 19 e = 5. 1 /(1.6 1 / ) = (b) J 5 I (5. 1 A) = = = 4 A π (. 1 m) 33 A/ m (c) (d) 7 f (4. 1 m/ s) a= = = m/ s Δx (.5 m) 17 ee ma kg m s a= = = ( )(1.6 1 / ) E 19 m e E = N / K = m = kg m s = ( )(4. 1 / ) J 5 19 Eery second, 5. 1 /1.6 1 = Thus, the total power deliered is, 14 electrons strike the screen. P J s = ( )( ) =.3W

7 Physics 1 / Summer 9 Name: ANSWER KEY h. 33 Quiz HOOSE ONE OF THE TWO PROBEMS. Problem 1: Using the Biot-Saart law, determine the magnetic field at the center of the circular arc (point P ) when the angle is φ = π /. Show that your answer is consistent with the expression for the field of a complete circular current loop when φ = π : B = μ I /R. a loi de Biot et de Saart: B μ I ds rˆ 4π r = φ R P In this problem, the straight parts of the current, will not contribute to the field at point P since ds is parallel or antiparallel to r. In all cases, r points from a current element to the point P. For the current arc, the cross product ds r, points into the page, so this is the direction of the B-field at the point P. Also, the magnitude of r is just the radius of the arc, r r = R. Since ds is perpendicular to r o for the current arc, the angle between the two is 9 and the sine is 1. So we hae, φ μ I ds rˆ μ I ds μ I ds B = = = 4π 4π 4π R φ φ r r Now ds is an infinitesimal element of the arc, ds = Rdφ, so we may now integrate to find the field, π / I Rd I B = μ I 4 = μ φ μ 4πR = 8R π / φ π R, into the page. A complete circle would gie 4 this amount, or μ I /R.

8 Problem : A wire bent at a 9 o angle carries a 1 A current in the direction shown. The length of each segment is 5 cm. If a uniform magnetic field of.1 T is applied in the direction shown, what is the magnitude and direction of the net force on the wire? F = I l B onsider each segment separately. For the ertical segment, the force is into the page. The magnitude is, Fert = Il Bsin 45 o =.35 N The force on the horizontal segment is also into the page. The magnitude is the same, I 45 o B Fhoriz = Il Bsin 45 o =.35 N Thus, the net force is, F net =.71 N

9 Physics 1 / Summer 9 Name: ANSWER KEY h. 34 Quiz In the figure shown, a 5 cm long conducting bar is in contact with conducting rails and forms a closed circuit. The magnetic field is uniform, haing a magnitude of.65 T. (a) If the circuit resistance is 5. Ω, and the position of the bar is gien by x =.5t, use Faraday s law to determine the magnitude and direction of the current flowing in the bar at t =. s. (b) What is the force on the bar at t =. s? x We choose the normal to the loop area to be into the page, in the same direction of the field. This fixes the positie orientation of the loop to be clockwise. dφm d( Blx) ε = = dt dt d(.5 t ) ε = Bl =Blt dt ε ( t =. s) = (.65 T)(.5 m)(. s) =.35V.35V It ( =. s) = =.65 A 5. Ω The minus sign of our answer indicates the current flows counterclockwise; that is, the current flows up the page through the conducting bar. F = Il B= (.65 A)(.5 m)(.65 T) ( iˆ ) The force is gien by: F =.11iˆ N

10 Physics 1 / Summer 9 Name: ANSWER KEY h. 35 Quiz A series R circuit is connected to an A oltage of amplitude ε = 4V and frequency 44 Hz. The phase angle at this frequency is 3 o. If = 1 mh and = 33 μf, what is the circuit resistance, R? (b) What is the impedance of the circuit? (c) Determine the RMS current in the circuit, I rms. (d) alculate the RMS oltage across the capacitor, V ( rms ). X X (a) tanφ = R = π f= 33. Ω, X = 1/ π f = 11. Ω X X X R = = 38.5 Ω tanφ (b) ( ) Z X X R = + = 44.4 Ω (c) I rms ε rms 4 V = = = Z (44.4 Ω).38 A (d) V ( ) = I X = (.38 A)(11 Ω ) = 4.V rms rms

Physics 212. Motional EMF

Physics 212. Motional EMF Physics 212 ecture 16 Motional EMF Conductors moing in field nduced emf!! Physics 212 ecture 16, Slide 1 The ig dea When a conductor moes through a region containg a magnetic field: Magnetic forces may