uppose a point charge Q. , traveling with a constant velocity -12a, -sa, ms with
ID: 2266138 • Letter: U
Question
uppose a point charge Q. , traveling with a constant velocity -12a, -sa, ms with the charge 5 pts (i) How much where there is a magnetic field h-a+9a, Tesla. Note that xperiences a Lorentz magnetic force. is the magnetic force F on the charge at the point of entry? ma 5 pts (ii) f the region also has an electric field E present, how m electromagne I uch is the total agnetic force For exerted on the charge at the point of entry? at value of the electric field E would make the net force For disappear? Show your work. E= 3 pts (iv) What is the physical effect on the moving charge of the electromagnetic field described in (iii)? Explain in some detail. 2, Note that a power flow vector, called the Poynting vector S , measured in watts/m s given by E × , where . represents complex conjugate. If the region is nonmagnetic, find the Poynting vector S. Which direction is the power flowing? 6 pts (v) Explain.Explanation / Answer
Answer:-i) B = 7ax + 9ay , u = 12ay - 8az . The force one a moving charged particle is-
F = Q(u X B), where X is the vector cross product.
Thus F = Q( (12ay - 8az ) x ( 7ax + 9ay) ) = Q( 72ax - 56ay - 84az ) N
ii) Force due to electric field is Fe = QE, so total electromagnetic force is-
Ftot = Fe + Fm = QE + Q( 72ax - 56ay - 84az ) N
iii) For Ftot = 0, i.e QE = - Q( 72ax - 56ay - 84az )
Thus E = - ( 72ax - 56ay - 84az ) V/m = ( -72ax + 56ay + 84az ) V/m (Assuming Q in C)
iv) Since net force acting on the charged particle is zero, the particle keeps on moving with its velocity and direction. There will be no effect of either field on the charged particle, no deflection of the charged particle.
v) For non-magnetic materials H = B = 7ax + 9ay ,since no imaginary term is here so, H* = H = 7ax + 9ay i.e same.
Hence S = E x H* = ( -756ax + 588ay - 1040az ) Wm-2 . The direction of power flow is perpendicular to both vectors E and H.
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