There is a constant homogeneous electric field of `100 Vm^-1` within the region `x=0 and x=0.167 m` pointing in x-direction. There is a constant homogeneous mangetic field B within the region `x=0.167 m and x=0.334 m` pointing in the z-direction. A proton at rest at the origin is released in positive x-direction. Find the minimum strength of the magnetic field B, so that the proton is detected back at `x=0, y=0.167 m.` (mass of proton `= 1.67 xx 10^(-27) kg)`

Explanation

First of all the proton is accelerated in the electric field.
Then it enters in magnetic field and describes a circular path. After that it leaves the magnetic field and describes a circular path.
After that it leaves the magnetic field in negative direction. Its
motion is retarded in electric field. Finally, it strikes y-axis at
the same distance 0.167m.

Let us first calculate the velocity of the proton when it entres in
magnetic field after traversing in electric field.
Force acting on proton in electric field =eE
`:.` Acceleration of proton `a=((eE)/m)`
Using the formula `v^2+u^2=2as`, we have
`v^2=2((eE)/m)s=2((eE)/m) (0.167)....(i)`
Now consider the motion in magnetic field. The proton describes
a circular path of radius `(AC//2)`.
Hence, `(mv^2)/r=evB or B=(mv)/(er)....(ii)`
`:. B=m/(e(0.1667//2))xxsqrt([(2eE)/m(0.167)])`
`=2sqrt(((2m)/(e)xxE/(0.167)))=2sqrt([(2(2.67xx10^-27)(100))/(1.6xx10^(-19)(0.167))])`
`=(10^-2)/sqrt2=7.07xx10^-3T`