Answer:
Magnetic field strength required for this is 0.25 T
Explanation:
As we know that the proton moves in circular path in uniform magnetic field
so the radius of the path of the circle is given as
[tex]R = \frac{mv}{qB}[/tex]
here we know that
[tex]q = 1.6 \times 10^{-19}C[/tex]
[tex]m = 1.6 \times 10^{-27} kg[/tex]
[tex]R = 0.0200 m[/tex]
[tex]v = 5 \times 10^5 m/s[/tex]
now we have
[tex]B = \frac{1.6 \times 10^{-27} (5 \times 10^5)}{(1.6 \times 10^{-19})(0.02)}[/tex]
so we have
[tex]B = 0.25 T[/tex]
