Question

Chapter 21, Problem 07 your answer is partially correct. Try again. A magnetic field has a magnitude of 0.0012 T, and an electric field has a magnitude of 4.9 x 103 N/C. Both fields point in the same direction. A positive 2.1- C charge moves at a speed of 4.4 x 106 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge. BA Ē 9 90° Number Units N the tolerance is +/-2%

Answer 1

Problem of The magnetic The force on the charge due to magnetic field is Fm qUB singoo +(2-13106 @x 4.4x106 0.0012 x1) N Х le. out wards to the 20.0111N, This force acting along perpendicular to both i and on plane of paper. The force charge due to eleclue field Fe=qE 2.1x10-6 x 419 x103 AN 21.029 410-5 N This force acting along the direction of electrice field Ê so, Fm and fe are perpendicular to each other. Therefore the force the charge Fm + Fe2 nel . = .01)2 + (1.029 710-5J? N 00111 N

Question of -- For * on * * The e magnetic force charged particle moving in with relocity us in a magnetic field Bris E qWXBT a is charge of to the particle. Where Given t So, magnitude of force is Fmz ar B sino [o is angle between to and in magnetic Griven . q = 1.6 1.6x1019.c 5.38106 mis 800x1014 N: 0 = 90°.

There fore 8.0x1614's 1.681019 x 5.3% 10%*B x singoo B 80x1074 1.6 x 10 19 x 5.3x10 xsingoo 6 0.09 43 T D Some criven is perpendicular to the motion of particle. so, o direction of B is efther tnword berpendicular to the plane of paper. If the direction of B is perpendicularly of ng outward to the plane of paper then the direction of Fm will be as shown in given. Hence B is perpendicularly outward to the plane of paper. Problem 14 The four time require to for the particle to complete revolution [Bis magnetic field mis man of particle q is charge on particle] Ba 2x 3114 0.70x 5.7x108 1.57.x10&s one 271 m a B 271 S 2