A region of space has an electric field of 905 N/C in the +x direction, and a ma
ID: 2004045 • Letter: A
Question
A region of space has an electric field of 905 N/C in the +x direction, and a magnetic field of 1.60 T in the +z direction. (a) If a 2.75 C charged particle moving in the x-y plane experiences a net force of 4.35 x 10-3 N in the +x direction, what is the speed and direction of the charged particle? Include a diagram.
***My specific question is how to find the angle in the equation F=qVBsin. I know that theta is the angle between the direction of motion and the magnetic field but I don't know how to find with the given information above. Any help would be appreciated! Thanks!
Explanation / Answer
Given:
A region of space has an electric field of= E = 905 N/C
(along the +x direction)
A region of space has a magnetic field of = B = 1.60 T
(along the +z direction)
charge of the moving particle in this region = q = 2.75 C
= 2.75 x10-6 C
when particle moving in the x-y plane ,
it experiences a net force of = F_net = 4.35 x 10-3 N
(along the +x direction)
since ,particle moving in( xy -plane ) and magnetic field
applied along the (+z direction ) , angle between these is
= 90o
when the particle moves in the magnetic field
the amount of magnetic force exerting on the particl is
given by the mathematical fornula as
F_mag = q(vx B)
= q v B sin (here v is the speed of the charged particle)
lly, when the particle is moving in the electrical field
the force acting on the particle by the elevtric field is given as
F_ ele = E q
thus, Net force acting on the particle
F_net = F_mag + F_ele
= qvB sin + E q
by the given values shown in above
4.35 x 10-3 = (2.75 x10-6)( v )( 1.60) sin 90 + ( 905) (2.75 x10-6)
= 423 m/s
speed and direction of the charged particle = 423 m/s
direction: opp. to the plane
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.