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Unformatted text preview: CHAPTER 29 THE MAGNETIC FIELD ActivPhysics can help with these problems: Activities 13.4, 13.6, 13.7, 13.8 Section 29 2: The Magnetic Force and Moving Charge Problem 1. (a) What is the minimum magnetic field needed to exert a 5 10 15 .4 ×N force on an electron moving at 2 1 10 7 . ? × m/s (b) What magnetic field strength would be required if the field were at 45 ° to the electron’s velocity? Solution (a) From Equation 291b, B F e = = v sin , q which is a minimum when sin q = 1 (the magnetic field perpendicular to the velocity). Thus, B min ( .4 ) ( . )( . ) . . = × × × = × = 5 10 1 6 10 2 1 10 1 61 10 16 1 15 19 7 3 N C m/s T G. = (b) For q = ° 45 , B B B = ° = = min min sin . = 45 2 22 7 G. Problem 2. An electron moving at right angles to a 0.10T magnetic field experiences an acceleration of 6 0 10 15 . . × m/s 2 (a) What is the electron’s speed? (b) By how much does its speed change in 1 10 9 ns ( s = ) ? Solution (a) If the magnetic force is the only one of significance acting in this problem, then F ma e B = = v sin . q Thus, v = = = = × × × ° = × ma eB sin ( . )( ) ( . )( . ) sin .42 q 9 11 10 6 10 1 6 10 0 1 90 3 10 31 15 19 5 kg m/s C T m/s. 2 (b) Since F v B » × is perpendicular to v , the magnetic force on a charged particle changes its direction, but not its speed. Problem 3. What is the magnitude of the magnetic force on a proton moving at 2 5 10 5 . × m/s (a) at right angles; (b) at 30 ° ; (c) parallel to a magnetic field of 0 50 . T? Solution From Equation 291b, F e B F = = ° = × × = × v sin , , ( . )( . )( . ) . , q q so (a) when C m/s T N 90 1 6 10 2 5 10 0 5 2 0 10 19 5 14 (b) F F e B = × ° = × = ° = ( . ) sin . , sin . 2 0 10 30 1 0 10 14 14 N N and (c) v Problem 4. A magnetic field of 0 10 . T points in the x direction. A charged particle carrying 1 0 . C m enters the field region moving at 20 m/s. What are the magnitude and direction of the force on the particle when it first enters the field region if it does so moving (a) along the x axis; (b) along the y axis; (c) along the z axis; (d) at 45 ° to both x and y axes? Solution The magnetic force is q v B v v v × = × = × ( )( )( . ) $ $ , $ 1 20 0 1 2 C m/s T N where m m î î is a unit vector in the direction of the velocity of the particle when it first enters the field region. (a) If $ , $ . ( ) $ $ , v v v = × = = î î b For j the direction of the force is $ $ j × =  î k (along the negative z axis), while (c) $ $ . $ (cos sin $ ) $ , k î î î î k × = × = ° + ° × =  j j (d) v 45 45 2 = so the force is ( ) $ . 2 N m k Problem 5. A particle carrying a 50 C m charge moves with velocity v = + 5 0 3 2 . . $ î k m/s through a uniform magnetic field 684 CHAPTER 29 B = + 9 6 7 .4 ....
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This note was uploaded on 10/19/2010 for the course PHY 78647 taught by Professor Miller during the Spring '09 term at Albany College of Pharmacy and Health Sciences.
 Spring '09
 Miller
 Physics, Force

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