Version 001 HW#5 - Magnetism arts (00224) 1

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1 Version 001 HW#5 - Magnetism arts (004) 1 This print-out should hae 11 questions. Multiple-choice questions may continue on the net column or page find all choices before answering. Charged Particle in a Field points A charged particle is projected with its initial elocity parallel to a uniform magnetic field. What is the resulting path? 1. straight line parallel to the field. correct. straight line perpendicular to the field. 3. circular arc. 4. spiral. 5. parabolic arc. The force on a moing charge due to a magnetic field is gien by F = q B. If and B are parallel, then B = 0. Hence the force on the particle is ero, and the particle continues to moe in a straight line parallel to the field. Field Direction points Thedirectionofthemagneticfieldinacertain region of space is determined by firing a test charge into the region with its elocity in arious directions in different trials. The field direction is 1. the direction of the elocity when the magnetic force is a maimum.. perpendicular to the elocity when the magnetic force is ero. 3. None of these correct 4. the direction of the magnetic force. 5. one of the direction of the elocity when the magnetic field is ero. The magnetic field direction can be determinedaccordingtothepathofachargewhich moes in the field. Since F = q B, the component of elocity along the magnetic field doesn t contribute to the magnetic force; i.e., a charge moing along the magnetic field will keep its motion status. On the contrary, if the initial elocity in perpendicular to the magnetic force, there will be a centripetal force so that the charge will undergo circular motion. Proton in a Magnetic Field points A proton moing at m/s through a magnetic field of.4 T eperiences a magnetic force of magnitude N. What is the angle between the proton s elocity and the field? The charge on a proton is C and its mass is kg. Correct answer: Let : E = N, B =.4 T, = m/s, q p = C, and m p = kg. The Lorent force acting on a moing charged particle in a magnetic field is F = qb sinθ ( ) F θ = arcsin qb [ N = arcsin C

2 = Version 001 HW#5 - Magnetism arts (004) ] 1 ( m/s)(.4 T) Alpha Particle points Analphaparticlehasamassof kg and is accelerated by a oltage of 0.59 kv. Thechargeonaprotonis C. If a uniform magnetic field of T is maintained on the alpha particle and perpendicular to its elocity, what will be particle s radius of curature? i 1 i Find the direction of the magnetic force on i due to the magnetic field of i to the left. to the right correct Correct answer: m. Let : B = T, V = 0.59 kv = 59 V, m = kg, and q = e = C. From Newton s second law, The kinetic energy is F = qb = m r = qbr m. K = 1 m = q B r m = qv, so the particle s radius of curature is r = 1 V m B q 1 (59 V)(6.610 = 7 kg) T C = m. Magnetic Force Directions points Two parallel wires carry opposite current as shown. 3. out of the paper 4. to the right and downward 5. to the left and upward 6. into the paper 7. to the left and downward 8. to the right and upward i 1 i Wire 1, which carries a current i 1, sets up a magneticfield B 1,whichpointsintothepaper at the position of wire. The direction of B 1 is perpendicular to the wire, as illustrated below. The magnetic force on a length l of wire is F = i l B1. Since i flows downward, l B 1 is to the right. Parallel Wires points Two parallel wires carry equal currents in B 1 F

3 Version 001 HW#5 - Magnetism arts (004) 3 the opposite directions. Point A is midway between the wires, and B is an equal distance on the other side of the wires. A What is the ratio of the magnitude of the magnetic field at point A to that at point B? 1. B A = 3. B A = B A = 0 4. B A = 4 5. B A = 6. B A = 3 correct 7. B A = 5 8. B A = 1 9. B A = B A = 1 3 The magnetic field due to a long wire is B = µ 0I πr. Let the distance between the wires be r. The magnetic field at A due to the upgoing wire is B up,a = µ 0I π(r/) = µ 0I πr. The right-hand rule tells us the direction is into the paper. Due to the fact that A is B the same distance from both wires, the other wire gies a magnetic field at A of the same magnitude, also directed into the paper due to the right-hand rule. The total magnetic field at A is B A = B up,a = µ 0I πr Now, the field at B due to the upgoing wire is B up,b = again into the paper, while µ 0 I π(3r/) = µ 0I 3πr, B down,b = µ 0I π(r/) = µ 0I πr out of the paper. So = B down,b B up,b = µ ( 0I 1 1 ) πr 3 = µ 0I 3πr, out of the paper. Comparing magnitudes, we find µ 0 I B A = πr = 3. µ 0 I 3πr Long Solenoid points What current is required in the windings of a long solenoid that has 150 turns uniformly distributed oer a length of m in order to produce inside the solenoid a magnetic field ofmagnitude T? Thepermeablity of free space is Tm/A. Correct answer:.6 ma. Let : N = 150, l = m, and B = T..

4 Version 001 HW#5 - Magnetism arts (004) 4 Magnetic field inside the solenoid is keywords: B = µ 0N I l I = Bl µ 0 N, then = ( T)(0.578 m) µ 0 (150) =.6 ma. Particle Deflection points A negatiely charged particle moing parallel to the -ais enters a magnetic field (pointing intoofthepage),asshowninthefigurebelow. y q Figure: î is in the -direction, ĵ is in the y-direction, and ˆk is in the -direction. 1. F = ĵ correct. F = 0; no deflection 3. F = ˆk 4. F = +ĵ 5. F = +î 6. F = +ˆk 7. F = î Basic Concepts: Magnetic Force on a Charged Particle: F = q B Right-hand rule for cross-products. F F F ; i.e., a unit ector in the F direction. Solution: The force is F = q B. ( ) = B ˆk, = (+î), and q < 0, therefore, F = q B = q B [ (+î) = q B (+ĵ) F = ĵ. ( )] ˆk This is the third of eight ersions of the problem. Particle Deflection points A positiely charged particle moing paralleltothe-aisentersamagneticfield(pointingtowardthebottomofthepage),asshown. î is in the -direction, ĵ is in the y-direction, and ˆk is in the -direction. y +q 1. F = +ĵ correct. F = ˆk 3. F = +ˆk

5 Version 001 HW#5 - Magnetism arts (004) 5 4. F = 0; no deflection 5. F = ĵ 6. F = î 7. F = +î The force is F = q B, B = B = 1 ( ) ˆk +î, and q > 0, so F = + q B [( ) = + q B ˆk = + q B (+ĵ) F = +ĵ. ] ( î) ( ) ˆk, Particle Deflection points A negatiely charged particle moing parallel to the y-ais enters a magnetic field (pointing toward the right-hand side of the page), as shown. î is in the -direction, ĵ is in the y-direction, and ˆk is in the -direction. q 1. F = î. F = +ˆk 3. F = +ĵ 4. F = ˆk 5. F = +î y 6. F = 0; no deflection correct 7. F = ĵ The force is F = q B, B = B (+ĵ), = 1 ( +ĵ+ ˆk ), and q < 0, so F = q B = q B [(+ĵ)(+ĵ)] = q B (0) F = 0 no deflection. Particle Deflection points A negatiely charged particle moing at 45 angles to both the -ais and -ais enters a magnetic field (pointing towards the righthand side of the page), as shown in the figure. î is in the -direction, ĵ is in the y-direction, and ˆk is in the -direction. q 1. F = 1 ) ( ˆk +î. F = 1 ) (+ĵ ˆk 3. F = 1 ) ( ĵ+ˆk 4. F = 1 ) (+ˆk î 5. F = 1 ) (+ĵ+ˆk 6. F = 0; no deflection y

6 Version 001 HW#5 - Magnetism arts (004) 6 7. F = +ĵ correct 8. F = 1 (+ĵ î) 9. F = ĵ 10. F = 1 ( ĵ+î) The force is F = q B, B = B = 1 ( ) ˆk +î, and q < 0, so ( ) +ˆk, F = q B 1 = q B = q F = +ĵ. 1 B (+ĵ) [( ) ( )] ˆk +î +ˆk