Question

Find the direction & magnitude of the force exerted by magnetic fields on the moving charge a I=0.5A21 1 zom > N = lomis . =2nc zem 1=2421 2m I- 2 Ac

Answer 1

Solution: 1,-0.5A - razem a = 2nc uolomis r2= 20m 1, - 2A 2tr, 277r Magnetic field at A by I, = lo II ( outwards) Magnetic field at A by I = Mo Iz ( into the page) ③ Net magnetic field at A = B, B₂ s elo (1,-1, ) Tr - 409 x 10-7 (0.5 - 2) 211 x 2x109 = -1.5 x 10-5T o indictates into the page ie ③ Therefore at A ® F' Force on chare particle now is F = q vB sind 0 = 90° ( angle blw v8 B). F = av vB - 2x10-9 X 10 x 15 x10-5 F = 3 X 10-3 N direction of force will be given by right hand Thumb rule or cross product of Yx r which will be up ie in ty direction."