A particle of mass `m = 1.6 xx10^(-27)` kg and charge `q=1.6xx10^(-19)C` enters a region of a uniform magnetic field of flux density 1 T. The speed of the particle is `10^(5)` m/s.

The particle leaves the field at B. Assuming that magnetic field is directed along the inward normal in the plane of the paper, find AB length and `theta`.

Explanation

As the particle enters the magnetic field, it will travel in a circular path. The centre will be on the line perpendicular to its velocity and the radius r will be, `r=(mv)/(qB)`.

Direction of force will from`F=q(v xx B)` shows that centre will be outside the field as shown in figure.
Now, `angleEBO=90^(@)`
(EB is tangent to OB) and `angleEBC=45^(@)`, so `angleOBA=45^(@), angleOBA=angleOAB=45^(@)" " (OB=OA=r)`
So, `angleDAF=theta=45^(@)`
In `DeltaAOB, AB=2(OB) cos angleOBA`
`=2r cos 45^(@)`
`=2xx (mv)/(qB) *(1)/(sqrt(2))`
`=(2xx1.6 xx 10^(-27) xx 10^(5))/(1.6 xx 10^(-19) xx 1) xx (1)/(sqrt(2))`
`=1.41 xx 10^(-3) m`