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# chap26 - MAGNETISM FORCE AND FIELD 26 EXERCISES Section...

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MAGNETISM: FORCE AND FIELD 26 E XERCISES Section 26.2 Magnetic Force and Field 17. I NTERPRET This problem is about the magnetic force exerted on a moving electron. D EVELOP The magnetic force on a charge q moving with velocity v G is given by Equation 26.1: . B F qv B = × G G G The magnitude of B F G is | | | | | | sin B B F F qv B q vB θ = = × = G G G E VALUATE (a) The magnetic field is a minimum when sin 1 θ = (the magnetic field perpendicular to the velocity). Thus, 15 3 min 19 7 5.4 10 N 1.61 10 T 16.1 G | | (1.6 10 C)(2.1 10 m/s) B F B q v × = = = × = × × (b) For 45 , θ = ° the magnetic field is 15 min 19 7 5.4 10 N 2 22.7 G | | sin (1.6 10 C)(2.1 10 m/s)sin 45 B F B B q v θ × = = = = × × ° A SSESS The magnetic force on the electron is very tiny. The magnetic field required to produce this force can be compared to the Earth’s magnetic field, which is about 1 G. 18. (a) If the magnetic force is the only one of significance acting in this problem, then sin . F ma ev B θ = = Thus, 31 15 2 19 5 v / sin (9.11 10 kg)(6 10 m/s )/(1.6 10 C)(0.1 T)sin 90 3.42 10 m/s. ma eB θ = = × × × ° = × (b) Since F v B × G G G z is perpendicular to the magnetic force on a charged particle changes its direction, but not its speed. , v G 19. I NTERPRET In this problem we are asked to find the magnetic force exerted on a moving proton. D EVELOP The magnetic force on a charge q moving with velocity v G is given by Equation 26.1: . B F qv B = × G G G The magnitude of B F G is | | | | | | sin B B F F qv B q vB θ = = × = G G G The charge of the proton is 19 1.6 10 C. q = × E VALUATE (a) When 90 , θ = ° the magnitude of the magnetic force is 19 5 14 sin 90 (1.6 10 C)(2.5 10 m/s)(0.5 T) 2.0 10 N B F qvB = ° = × × = × (b) When 30 , θ = ° the force is 19 5 14 sin 30 (1.6 10 C)(2.5 10 m/s)(0.5 T)sin30 1.0 10 N B F qvB = ° = × × ° = × (c) When 0 , θ = ° the force is sin 0 0. B F qvB = ° = ) A SSESS The magnetic force is a maximum ,max ( | | B F q vB = when 90 θ = ° and a minimum when ,min ( 0 B F = ) 0 . θ = ° 20. An electron, moving perpendicularly to the Earth’s magnetic field near the surface, experiences a maximum magnetic force of 19 4 3 19 31 v 2 / (1.6 10 C)(10 T) 2(10 eV)(1.6 10 J/eV)/(9.11 10 kg) e B eB K m = × × × . . 16 3 10 N. × The weight of an electron near the Earth’s surface is 31 2 30 (9.11 10 kg)(9.8 m/s ) 9 10 N, mg = × × . a factor of nearly times smaller. 14 3 10 × 21. I NTERPRET This problem is about the speed of a given charge if it is to pass through the velocity selector undeflected. 26.1

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