This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 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 291b, 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 0.10T 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 . 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 291b, 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 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 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 . 2 CHAPTER 29 Solution (a) From Equation 272, 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 . . ) ( . ....
View
Full
Document
This note was uploaded on 05/12/2008 for the course PHYS 299 taught by Professor Hoston,amahd,bakanowski during the Spring '08 term at University of Louisville.
 Spring '08
 Hoston,Amahd,Bakanowski
 Force

Click to edit the document details