Question

A-4.90 LC charge is moving at a constant speed of 6.90x105 m/s in the to-direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic field vector it produces at the following points.

Answer 1

Use the equation for magnetic field due to a moving charged particle at a given point to find the required solution as shown below

11 s q= charge ? = velocity Here 2 = 0.5mi V = 6.9x105 mis so, řxř = (6-9x109 €)x(0-5) = 0 So, B=0 ş) Bar, By, B2 = 10,0,01 T Here ² = 0.5mj o. r=1% 1= 0.5m so, B = Mo a CWxr) 41 73 = lo 64.94c)(6.9x105 î x 0.55) 10.5m3 = -1.35x10-6 T Ê So, Bx, By, B2 - 0, 0, -1.35x10-6) T c) 705 +0.5 r = 20 l = 0.552 = 0.707m Be the acUxy?) = Mo (4.94c)(6.9x105 î x(0-sî+0.5.j')) (0.707 m)3 = -4.78X 10-7 T Bx, By, By = 0, 0, -4.78X107) 7 D) Here 8=0.5mk , r=0.5m B = Mo afvxr?) 1 40 3 = Ho (4.quc)(6.9x105 î x0.512) (0.5m)? = 1:35x10-6 T 3) Ba, By, B, = 10, 1.35x10-6, 0 T