-left. -out of the paper. -down. -into the paper. -right. -up. -not at all. The
ID: 1260493 • Letter: #
Question
-left.
-out of the paper.
-down.
-into the paper.
-right.
-up.
-not at all.
The shape of its trajectory will be ...??????
-straight, continuing at same speed.
-parabolic, curving out of the paper.
-parabolic, curving up.
-circular, clockwise.
-straight, slowing down.
-straight, speeding up.
-circular, counter-clockwise.
-parabolic, curving down.
-parabolic, curving into the paper.
If the field is 5 mT, the mass of the particle is 49?10?9 kg, its initial speed 8.4 m/s, and its charge -6 ?C, what is the magnitude of its acceleration?
What is the speed of the particle 1.1 seconds later?
A negatively charged particle is moving with a velocity v in a large homogeneous magnetic field as indicated. It will accelerate ...?????? -left. -out of the paper. -down. -into the paper. -right. -up. -not at all. The shape of its trajectory will be ...?????? -straight, continuing at same speed. -parabolic, curving out of the paper. -parabolic, curving up. -circular, clockwise. -straight, slowing down. -straight, speeding up. -circular, counter-clockwise. -parabolic, curving down. -parabolic, curving into the paper. If the field is 5 mT, the mass of the particle is 49½10^?9 kg, its initial speed 8.4 m/s, and its charge -6 ?C, what is the magnitude of its acceleration? What is the speed of the particle 1.1 seconds later?Explanation / Answer
Using the right hand rule it will accelerate downward.
It will go in a circular counterclockwise motion (since force will always be perpendicular to it's motion)
F= q* VxB= 2.52 *10^(-7)N
F=ma so 2.52*10^(-7)= a *49*10^(-9)
a= 5.1429 m/s^2
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