# ch28 - CHAPTER The Magnetic Field 28 1 When a cathode-ray...

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CHAPTER 28 The Magnetic Field 1* · When a cathode-ray tube is placed horizontally in a magnetic field that is directed vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube in Figure 28-30. The correct path is ____. ( a ) 1 ( b ) 2 ( c ) 3 ( d ) 4 ( e ) 5 ( b ) 2 · Why not define B to be in the direction of F , as we do for E ? One cannot define the direction of the force by fiat. By experiment, F is perpendicular to B . 3 · Find the magnetic force on a proton moving with velocity 4.46 Mm/s in the positive x direction in a magnetic field of 1.75 T in the positive z direction. Use Equ. 28-1 F = 1.25 pN i · k = –1.25 pN j 4 · A charge q = – 3.64 nC moves with a velocity of 2.75 × 10 6 m/s i . Find the force on the charge if the magnetic field is ( a ) B = 0.38 T j , ( b ) B = 0.75 T i + 0.75 T j , ( c ) B = 0.65 T i , ( d ) B = 0.75 T i + 0.75 T k . ( a ), ( b ), ( c ), ( d ) Use Equ. 28-1 ( a ) F = –3.8 mN i × j = -3.8 mN k ( b ) F = –7.5 mN k ( c ) F = 0 ( d ) F = 7.5 mN j 5* · A uniform magnetic field of magnitude 1.48 T is in the positive z direction. Find the force exerted by the field on a proton if the proton’s velocity is ( a ) v = 2.7 Mm/s i , ( b ) v = 3.7 Mm/s j , ( c ) v = 6.8 Mm/s k , and ( d ) v = 4.0 Mm/s i + 3.0 Mm/s j . ( a ), ( b ), ( c ), ( d ) Use Equ. 28-1 ( a ) F = 0.639 pN i × k = –0.639 pN j ( b ) F = 0.876 pN i ( c ) F = 0 ( d ) F = 0.71 pN i – 0.947 pN j 6 · An electron moves with a velocity of 2.75 Mm/s in the xy plane at an angle of 60 ° to the x axis and 30 ° to the y axis. A magnetic field of 0.85 T is in the positive y direction. Find the force on the electron. Use Equ. 28-1 F = –0.44(cos 60 o i + cos 30 o j ) × 0.85 j pN = –0.187 pN k 7 · A straight wire segment 2 m long makes an angle of 30 ° with a uniform magnetic field of 0.37 T. Find the magnitude of the force on the wire if it carries a current of 2.6 A. Use Equ. 28-4 F = BI l sin ? = 0.962 N 8 · A straight wire segment I l = (2.7 A)(3 cm i + 4 cm j ) is in a uniform magnetic field B = 1.3 T i . Find the

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Chapter 28 The Magnetic Field force on the wire. Use Equ. 28-4 F = –0.140 N k 9* · What is the force (magnitude and direction) on an electron with velocity v = (2 i – 3 j ) × 10 6 m/s in a magnetic field B = (0.8 i + 0.6 j – 0.4 k ) T? Use Equ. 28-4 F = –0.192 pN i 0.128 pN j 0.576 pN k ; F = 0.621 pN 10 ·· The wire segment in Figure 28-31 carries a current of 1.8 A from a to b . There is a magnetic field B = 1.2 T k . Find the total force on the wire and show that it is the same as if the wire were a straight segment from a to b . Use Equ. 28-4 If the wire is straight from a to b , l = 3 cm i + 4 cm j F = –0.0684 N j + 0.0864 N i ; F = I l · B = 0.0864 N i – 0.0684 N j 11 ·· A straight, stiff, horizontal wire of length 25 cm and mass 50 g is connected to a source of emf by light, flexible leads. A magnetic field of 1.33 T is horizontal and perpendicular to the wire. Find the current necessary
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## This note was uploaded on 05/01/2008 for the course PHYS 13/14 taught by Professor Millan/thorstensen during the Spring '08 term at Dartmouth.

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ch28 - CHAPTER The Magnetic Field 28 1 When a cathode-ray...

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