AP Physics Test Magnetic Fields; Sources of Magnetic Field

Transcription

1 AP Physics Test Magnetic Fields; Souces of Magnetic Field Pat I. Multiple hoice (4 points each) hoose the one best answe to each of the following poblems. axis 2 A = 0.05 T 0.3 m 0.3 m 1 (AP). A squae loop of wie 0.3 metes on a side caies a cuent of 2 ampees and is located in a unifom 0.05Tesla magnetic field. The left side of the loop is aligned along and attached to a fixed axis. When the plane of the loop is paallel to the magnetic field in the position shown above, what is the magnitude of the toque exeted on the loop about the axis? a N m b N m c N m d N m e. 111 N m 2 (AP). A lage paallelplate capacito is being chaged and the magnitude of the electic field between the plates of the capacito is inceasing at the ate de/dt. Which of the following statements is coect about the magnetic field in the egion between the plates of the chaging capacito? a. It is paallel to the electic field. b. Its magnitude is diectly popotional to de/dt. c. Its magnitude is invesely popotional to de/dt. d. Nothing about the field can be detemined unless the chaging cuent is known. e. Nothing about the field can be detemined unless the instantaneous electic field is known. v 3 (AP). A negatively chaged paticle in a unifom magnetic field moves with constant speed v in a cicula path of adius as shown above. Which of the following gaphs best epesents the adius as a function of the magnitude of if the speed v is constant? a. b. c. d. e.

2 I (into page) axis 4 (AP). I (out of page) A coss section of a long solenoid that caies cuent I is shown above. All of the following statements about the magnetic field inside the solenoid ae coect EXEPT: a. is diected to the left. b. An appoximate value fo the magnitude of may be detemined using Ampee s Law c. The magnitude of is popotional to the cuent I. d. The magnitude of is popotional to the numbe of tuns of wie pe unit length. e. The magnitude of is popotional to the distance fom the axis of the solenoid. 5 (AP). At point X a chaged paticle has a kinetic enegy of 9 micojoules (µj). It follows the path shown above fom X to Y though a egion in which thee is an electic field and a magnetic field. At Y, the paticle has a kinetic enegy of 11 µj. What is the wok done by the magnetic field on the paticle? a. 11 µj b. 2 µj c. 2 µj d. 11 µj e. None of the above 6. A 2.0 m wie caies a cuent of 15 A diected along the positive x axis in a egion whee the magnetic field is unifom and given by = (30i 40 j) mt. What is the esulting magnetic foce on the wie? a. (+1.2 k) N b. (1.2 k) N c. (1.5 k) N d. (+1.5 k) N e. (+0.90 k) N 7. An electon which moves though a velocity selecto (E = 4.0 kv/m, = 2.0 mt) subsequently follows a cicula path (adius = 4.00 mm) in a unifom magnetic field. What is the magnitude of this magnetic field? a. 1.8 mt b. 2.4 mt c. 3.2 mt d. 2.8 mt e. 4.6 mt P a 2a X Y 8 (AP). Two long paallel wies ae a distance 2a apat as shown above. Point P is in the plane of the wies and a distance a fom wie X. When thee is a cuent I in wie X and no cuent in wie Y, the magnitude of the magnetic field at P is 0. When thee ae equal cuents I in the same diection in both wies, the magnitude of the magnetic field at P is: a. (2/3) 0 b. 0 c. (10/9) 0 d. (4/3) 0 e A solenoid 4.0 cm in adius and 4.0 m in length has 8000 unifomly spaced tuns and caies a cuent of 5.0 A.

3 onside a plane cicula suface (adius = 2.0 cm) located at the cente of the solenoid with its axis coincident with the axis of the solenoid. What is the magnetic flux though this suface? (1 Wb = 1 T m 2 ) a. 63 µwb b. 16 µwb c mwb d. 10 µwb e. 5.0 µwb Pat II. Fee Response (20 points each) ε filament 10 (AP). An electon fom a hot filament in a cathode ay tube is acceleated though a potential diffeence ε. It then passes into a egion of unifom magnetic field, diected into the page as shown above. The mass of the electon is m and the chage has magnitude e. a. Find the potential diffeence necessay to give the electon a speed v as it entes the magnetic field. b. On the diagam above, sketch the path of the electon in the magnetic field. c. In tems of mass m, speed v, chage e, and field stength, develop an expession fo, the adius of the cicula path of the electon. d. An electic field E is now established in the same egion as the magnetic field, so that the electon passes though the egion undeflected. i. Detemine the magnitude of E. ii. Indicate the diection of E on the diagam above.

4 11 (AP). A long coaxial cable, a section of which is shown above, consists of a solid cylindical conducto of adius a, suounded by a hollow coaxial conducto of inne adius b and oute adius c. The two conductos each cay a unifomly distibuted cuent I, but in opposite diections. The cuent is to the ight in the oute cylinde and to the left in the inne cylinde. Assume µ = µo fo all mateials in this poblem. a. Use Ampee s law to detemine the magnitude of the magnetic field at a distance fom the axis of the cable in each of the following cases. i. 0 < < a ii. a < < b b. What is the magnitude of the magnetic field at a distance >= c fom the axis of the cable? c. On the axes below, sketch the gaph of the magnitude of the magnetic field as a function of, fo all values of. You should estimate and daw a easonable gaph fo the field between b and c athe than attempting to detemine an exact expession fo the field in this egion. a b c

5 q P oss section The coaxial cable continues to cay cuents I as peviously descibed. In the coss section above, cuent is diected out of the page towad the eade in the inne cylinde and into the page in the oute cylinde. Point P is located between the inne and oute cylindes, a distance fom the cente. A small positive chage Q is intoduced into the space between the conductos so that when it is at point P its velocity v is diected out of the page, pependicula to it, and paallel to the axis of the cable. d. i. Detemine the magnitude of the foce on the chage q at point P in tems of the given quantities. ii. Daw an aow on the diagam at P to indicate the diection of the foce. e. If the cuent in the oute cylinde wee evesed so that it is diected out of the page, how would you answes to (d) change, if at all? ρ 12 (AP). The solid nonconducting cylinde of adius R shown above is vey long. It contains a negative chage evenly distibuted thoughout the cylinde, with volume chage density ρ. Point is outside the cylinde at a distance 1 fom its cente and point is inside the cylinde at a distance 2 fom its cente. oth points ae in the same plane, which is pependicula to the axis of the cylinde. a. On the following cosssectional diagam, daw vectos to indicate the diections of the electic field at points and. 2 R

6 b. Using Gauss law, deive expessions fo the magnitude of the electic field E in tems of, R, ρ, and fundamental constants fo the following two cases: i. > R (outside the cylinde) ii. < R (inside the cylinde) I R 2 I Anothe cylinde of the same dimensions, but made of conducting mateial, caies a total cuent I paallel to the length of the cylinde, as shown in the diagam above. The cuent density is unifom thoughout the cosssectional aea of the cylinde. Points and ae in the same positions with espect to the cylinde as they wee fo the nonconducting cylinde. c. On the following cosssectional diagam in which the cuent is out of the plane of the page (towad the eade), daw vectos to indicate the diections of the magnetic field at points and. d. Use Ampee s law to deive an expession fo the magnetic field inside the cylinde in tems of, R, I, and fundamental constants.

7 ANSWERS/SOLUTIONS: 1. b; τ= x F = IL 2. b; Use AmpeeMaxwell Law 3. b; =mv/q indicates invese elationship between and 4. e 5. e 6. b; Use F=ILx, using i,j notation 7. d 8. d; use =µi/2πa 9. =µni, and Φ = da We did this in class! 11. a. i. Use ds=µi, and I/Io = π 2 /πa 2 to get = (µoi)/(2πa 2 ) ii. Use Ampee s Law to get =(µoi)/(2π) b. ecause thee is no net cuent flow, = 0 outside the cable. c. Diectly popotional up to a, invesely popotional to b, then deceasing moe (concave up) to 0 at c. d. i. F=qv x ; v & ae pependicula, to F=(qvµI)/(2π) ii. Towad cente e. Answe would not change extenal cuents don t affect the field. 12. a. E1 is down towad at P1, E is smalle, and towad at P2. b. i. E = qin/(ε2πl); then use ρ = qin/volume to get the qintenal. E = (ρ R 2 )/2εο) ii. E = (ρ )/2εο) c. at P1 points to the left; at P2 points up and to the ight d. = (µoi)/(2πr 2 )