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Unformatted text preview: MAGNETISM: FORCE AND FIELD 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 r is given by Equation 26.1: . B F qv B = r r r The magnitude of B F r is       sin B B F F qv B q vB = = = r r r E VALUATE (a) The magnetic field is a minimum when sin 1 = (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 , = 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  = = = = A SSESS The magnetic force on the electron is very tiny. The magnetic field required to produce this force can be compared to the Earths 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 = = Thus, 31 15 2 19 5 v / sin (9.11 10 kg)(6 10 m/s )/(1.6 10 C)(0.1 T)sin90 3.42 10 m/s. ma eB  = = = (b) Since F v B r r r z is perpendicular to , v r the magnetic force on a charged particle changes its direction, but not its speed. 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 r is given by Equation 26.1: . B F qv B = r r r The magnitude of B F r is       sin B B F F qv B q vB = = = r r r The charge of the proton is 19 1.6 10 C. q = E VALUATE (a) When 90 , = the magnitude of the magnetic force is 19 5 14 sin90 (1.6 10 C)(2.5 10 m/s)(0.5 T) 2.0 10 N B F qvB = = = (b) When 30 , = the force is 19 5 14 sin30 (1.6 10 C)(2.5 10 m/s)(0.5 T)sin30 1.0 10 N B F qvB = = = (c) When 0 , = the force is sin 0 0. B F qvB = = A SSESS The magnetic force is a maximum ,max (   ) B F q vB = when 90 = and a minimum ,min ( 0) B F = when 0 . = 20. An electron, moving perpendicularly to the Earths 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 Earths surface is 31 2 30 (9.11 10 kg)(9.8 m/s ) 9 10 N, mg = . a factor of nearly 14 3 10 times smaller. 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 26 26.2 Chapter 26 D EVELOP In the presence of both electric and magnetic fields, the force on a moving charge is (see Equation 26.2): ( ) E B F F F q E v B = + = + r r r r r r The condition for a charged particle to pass undeflected through the velocity selector is , E B F F =  r r or / . v E B = E VALUATE Substituting the values given in the problem statement, we obtain 24 kN/C 400 km/s 0.06 T0....
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 Spring '09
 Prof.Kim
 Physics, Charge, Magnetism, Force

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