A particle of mass 4.814 times 10-26 kg and charge of 4.8 times 10-19 C is accel

Question

A particle of mass 4.814 times 10-26 kg and charge of 4.8 times 10-19 C is accelerated from rest in the plane of the page through a potential difference of 180 V between two parallel plates as shown. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0.0993 T. The particle curves in a semicircular path and strikes a detector. What is the magnitude of the force exerted on the charged particle as it enters the region of the magnetic field B rightarrow ? Answer in units of N

Explanation / Answer

before going through the hole it experiences an electric force given by Eq or Vq/d. (d is the distance from one plate to another, also known as x)

knowing that a=F/m so it's acceleration is Vq/(d*m)

the final velocity as it enters the hole is given by: vf2=vi2+2ax

so v2=2(Vq/(x*m) * x = (2Vq/m)

v=(2Vq/m)

FB= qvxB= q*(2Vq/m)*B

there you go

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