A proton with speed v = 3 times 106 m/s moves at an angle as shown into a unifor

Question

A proton with speed v = 3 times 106 m/s moves at an angle as shown into a uniform magnetic field of strength B = 4 times 10-4 T in the +y direction. Determine the magnitude and direction of the magnetic force on the proton as it firsts enters the field. Calculate the orbital radius of the proton in the field. Describe the subsequent path of the proton as it travels through the field. Describe the motion of the proton as it leaves the magnetic field, assuming no fringe effects Now an electric field E is added to the same region, along with the magnetic field. What are the strength and direction of the electric field needed so the proton travels through both fields without being deflected.

Explanation / Answer

a)

Use right hand rule >>>> +z

b)

F = q v B sin = 1.6e-19*3e6*4e-4*sin(90+30) = 1.6627688e-16 N

m v2/R = F

>>>> R = m v2/F = 1.673e-27*(3e6*3e6)/1.6627688e-16 = 90.6 m

c)

The proton travels on a semi circle with radius R and then exits the magnetic field.

d)

the proton will travel on a straght line with speed of 3*10^6 m/s

e)

qE = q v B sin = 1.6627688e-16

>>> E = 1.6627688e-16/1.6e-19 = 1039 V/m

>>> direction: -z

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