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CH29 - CHAPTER 29 THE MAGNETIC FIELD ActivPhysics can help...

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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 21 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 29-1b, B F e = = v sin , θ which is a minimum when sin θ = 1 (the magnetic field perpendicular to the velocity). Thus, B min ( .4 ) ( . )( . ) . . = × × × = × = - - - 5 10 16 10 21 10 161 10 161 15 19 7 3 N C m/s T G. = (b) For θ = ° 45 , B B B = ° = = min min sin . = 45 2 22 7 G. Problem 2. An electron moving at right angles to a 0.10-T 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 . θ Thus, v = = = = × × × ° = × - - ma eB sin ( . )( ) ( . )( . ) sin .42 θ 911 10 6 10 16 10 01 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 29-1b, F e B F = = ° = × × = × - - v sin , , ( . )( . )( . ) . , θ θ so (a) when C m/s T N 90 16 10 2 5 10 0 5 2 0 10 19 5 14 (b) F F e B = × ° = × = ° = - - ( . ) sin . , sin . 2 0 10 30 10 10 0 0 14 14 N N and (c) v Problem Solution The magnetic force is q v B v v v × = × = × ( )( )( . ) , 1 20 01 2 C m/s T N where μ μ 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 0 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 μ k Problem 5. A particle carrying a 50- C μ charge moves with velocity v = + 5 0 3 2 . . î k m/s through a uniform magnetic field B = + 9 6 7 .4 . î j T. (a) What is the force on the particle? (b) Form the dot products F v F B and to show explicitly that the force is perpendicular to both v and B .
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2 CHAPTER 29 Solution (a) From Equation 27-2, F v B k = × = + × + = × × + × - q ( )( . ) ( .4 . ) ( )( . . .4 50 5 3 2 9 6 7 50 10 5 6 7 3 2 9 6 C m/s T N μ î î k j j 3 2 6 7 1072 1504 1675 10 3 . . ) ( . . . ) × = - + + × - î î k j N. (The magnitude and direction can be found from the components, if desired.) (b) The dot products F v F B and are, respectively, proportional to ( . )( ) ( . )( . ) , - + = 1 072 5 1675 3 2 0 and ( . )( .4) ( . )( . ) , - + = 1 072 9 1504 6 7 0 since the cross product of two vectors is perpendicular to each factor. (We did not round off the components of F , so that the vanishing of the dot products could be exactly confirmed.) Problem 6. Moving in the x direction, a particle carrying 1 0 . C μ experiences no force. Moving with speed v at 30 ° to the x axis, the particle experiences a magnetic force of 2 0 . N. What magnetic force would it experience if it moved along the y axis with speed v ?
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