Solution :
The relationship between the strength of magnetic field and the radiusof a charged particle's path is obtained through Newton's second law, which is given by :
F = ma
F = qvB and [tex]$a=\frac{v^2}{r}$[/tex]
Substituting these values in the second law of Newton,
[tex]$qvB=\frac{mv^2}{r}$[/tex]
Now solving for B, we get:
[tex]$B = \frac{mv}{rq}$[/tex]
[tex]$=\frac{(1.67 \times 10^{-27})(5 \times 10^{7})}{2\times 1.6 \times 10^{-19}}$[/tex]
= 0.261 T
The field strength can be obtained by using the technology of today.
