(a) What is the magnitude of the magnetic force on theparticle? F = N Now a unif

Question

(a) What is the magnitude of the magnetic force on theparticle?

F = N    

Now a uniform electric field is applied so that the net force onthis particle is zero.

At the moment the particle first enters the region of themagnetic field, what direction is the force due to the magneticfield on the particle? Answer by giving a numerical code accordingto the following scheme:   (1) =+x,   (2) = -x,   (3) = +y ,   (4) =-y ,   (5) = +z ,   (6) = -z

Direction of force due to magnetic field =     

(b) What are the components of the electric field?

Ex = N/C   

Ey = N/C   

Ez = N/C    

(c) If the incoming particle were replaced by one with theopposite electric charge but the same initial velocity,what electric field would be required to ensure a straight linemotion? (Magnetic field is unchanged.)

Ex' = N/C   

Ey' = N/C   

Ez' = N/C    

A charged particle (q = +2.7 µC) moves at speedv0 = 51 m/s in the +x-direction. Atx = 0 it enters a region where a constant magnetic fieldB = 3.5 T is directed in the +z direction asshown in the diagram below. (The y-axis points into thescreen.)

Explanation / Answer

a1. F=qvb a2.using right hand rule -y b. F=qE    E=qvb/q=vb Ex=0 Ez=0 since the only force acting in on the y direction Ey=vb c. same as part b

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