F= qvBsin(theta) (-.16x10^-19 C)(6 x 10^6)(1T)(1) = -9.6 x 10^-13 N I also know
ID: 2075096 • Letter: F
Question
F= qvBsin(theta)
(-.16x10^-19 C)(6 x 10^6)(1T)(1) = -9.6 x 10^-13 N
I also know th solutions for all the Directions of F but do not understand how to solve for them! Can someone please explain in detail how to find the directions of each?
a) North [1st] / South [second] d) west g) east
b) east e) North h) south
c) east f)No B-field/east i)No force acting on this point
b) Can someone explain how to find the value and direction? i know the answer for b to e is:
value= zero sinQ = 0
direction is east
c) Answer: exit 2, but why is it exi 2 please explain?
11 10 WP E X g) X h) X (i) if) (e) Question 1. An E&M; classically trained (not a quantum mechanic particle) q-charge of value (-1.6x10 9C) enter a maze with velocity vector v (6 x 10 6 m/s in East direction. Calculate the force and direction at the locations (a (gl-0. Going from (b) to (e) What will the value and direction ofB-field have to be? Finally, at what exit (1-11) will q-charge escape from the maze, and learn the quantum mechanic way.Explanation / Answer
ans :-
the path of the electron is a -> g -> h-> d -> a -> b -> f -> e ->c -> 2
now we find the magnitude and direction of force according to the path
For a
the direction of velocity is east
the direc tion of magnetic field is out of plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = north
For g
the direction of velocity is north
the direc tion of magnetic field is in the plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = east
For h
the direction of velocity is east
the direc tion of magnetic field is into the plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = south
For d
the direction of velocity is south
the direc tion of magnetic field is into plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = west
For a
the direction of velocity is west
the direc tion of magnetic field is out of plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = south
For b
the direction of velocity is south
the direc tion of magnetic field is out of plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = east
For f
the direction of velocity is east
the direc tion of magnetic field is 0
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 0 direction = east same as b
For e
the direction of velocity is east
the direc tion of magnetic field is out of plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = north
For c
the direction of velocity is north
the direc tion of magnetic field is into plane of paper
Force = q (v x B) direction can be found out using right hand screw rule( the answer will be opposite since the charge is negative)
magnitude = 1.6 * 10-19 ( 6*106 *1) = 9.6 * 10-13 N direction = east
at 2
exit through 2
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