005 10.0 polnts A particle of muss 1.1288 × 10-25 kgand charge of 3.2 x 10 C is

Question

005 10.0 polnts A particle of muss 1.1288 × 10-25 kgand charge of 3.2 x 10 C is accelerated from rest in the plane of the page through a po tential difference of 287 V betweenm two paral lel plates as shown. The particle is injected through hole in the right-hand plate into a region of space containing a uniform mag- netic field of magnitude 0.218 T. The particle curves in a semicircular path and strikes detector. Regilon of tm Magnetic bole Field B What is the manitude of the force exerted , on the charged particleas it enters the region s of the inagnetic feld B of Answer in nits of N

Explanation / Answer

We need to equate the electric field to the Kinetic Energy (as the particle is just entering the field).

q V = 1/2 m v^2

V - Potential

v - velocity

So we have-:

3.2 x 10^-19 * 287 = 1/2 * 1.1288 x 10^-25 * v^2

v = 40338.75

Force = q v b (sin theta)

(you are not given an angle so assume sin theta = 90 -: thus = 1 and can be removed)

F = (3.2 x 10^-19)( 40338.75)(0.218)

= 2.814 x 10^-15 N

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