# Ch29 - CHAPTER Sources of the Magnetic Field 29 1 Compare...

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CHAPTER 29 Sources of the Magnetic Field 1* Compare the directions of the electric and magnetic forces between two positive charges, which move along parallel paths ( a ) in the same direction, and ( b ) in opposite directions. ( a ) The electric forces are repulsive; the magnetic forces are attractive (the two charges moving in the same direction act like two currents in the same direction). ( b ) The electric forces are again repulsive; the magnetic forces are also repulsive. 2 At time t = 0, a particle with charge q = 12 µ C is located at x = 0, y = 2 m; its velocity at that time is v = 30 m/s i . Find the magnetic field at ( a ) the origin; ( b ) x = 0, y = 1 m; ( c ) x = 0, y = 3 m; and ( d ) x = 0, y = 4 m. ( a ) Use Equ. 29-1; r = –2 m j , v = 30 m/s i ( b ) r = –1 m j ( c ) r = 1 m j ( d ) r = 2 m j B = 10 –7 q ( v × r )/ r 3 ; B = –9 × 10 –12 T k B = –3.6 × 10 –11 T k B = 3.6 × 10 –11 T k B = 9 × 10 –12 T k 3 For the particle in Problem 2, find the magnetic field at ( a ) x = 1 m, y = 3 m; ( b ) x = 2 m, y = 2 m; and ( c ) x = 2 m, y = 3 m. ( a ) Use Equ. 29-1; r = 1 m i + 1 m j ; v = 30 m/s i ( b ) r = 2 m i ( c ) r = 2 m i + 1 m j B = 10 –7 q ( v r )/ r 3 ; B = 1.27 × 10 –11 T k B = 0 B = 3.22 × 10 –12 T k 4 A proton (charge + e ) traveling with a velocity of v = 1 × 10 4 m/s i + 2 × 10 4 m/s j is located at x = 3 m, y = 4 m at some time t . Find the magnetic field at the following positions: ( a ) x = 2 m, y = 2 m; ( b ) x = 6 m, y = 4 m; and ( c ) x = 3 m, y = 6 m. ( a ) Use Equ. 29-1; q = 1.6 × 10 –19 C; r = (– i –2 j ) m; v = ( i + 2 j ) × 10 4 m/s; B = 10 –7 q ( v r )/ r 3 ; ( b ) r = 3 m i ( c ) r = 2 m j B = 0 ( v and r are colinear) B = –3.56 × 10 –23 T k B = 4 × 10 –23 T k 5* An electron orbits a proton at a radius of 5.29 × 10 –11 m. What is the magnetic field at the proton due to the orbital motion of the electron? 1. Determine the speed of the electron; mv 2 / r = ke 2 / r 2 2. Use Equ. 29-1; B = ( 0 e 2 /4 π r 2 ) r /m k ; evaluate v = r /m k e B = 12.5 T

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Chapter 29 Sources of the Magnetic Field B 6 ∙∙ Two equal charges q located at (0, 0, 0) and (0, b , 0) at time zero are moving with speed v in the positive x direction ( v << c ). Find the ratio of the magnitudes of the magnetic and electrostatic force on each. Note that v and r , where r is the vector from one charge to the other, are at right angles. The field B due to the charge at the origin at the location (0, b , 0) is perpendicular to v and r , i.e., in the z direction, and its magnitude is B = ( µ 0 /4 π ) qv / b 2 . The magnitude of the force on the moving charge at (0, b , 0) is F B = qvB = ( 0 /4 ) q 2 v 2 / b 2 . Applying the right hand rule the direction of the force is toward the charge at the origin; i.e., the magnetic interaction between the two moving charges is attractive. The electrostatic force is repulsive and of magnitude F E = (1/4 πε 0 ) q 2 / b 2 . Thus F B / F E = ε 0 0 v 2 = v 2 / c 2 , where c is the speed of light in vacuum. 7 The Biot-Savart law is similar to Coulomb's law in that both ( a ) are inverse square laws. ( b ) deal with forces on charged particles.
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## This note was uploaded on 04/23/2008 for the course PHYS 270 taught by Professor Drake during the Spring '08 term at Maryland.

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Ch29 - CHAPTER Sources of the Magnetic Field 29 1 Compare...

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