Answer:
0.253 m
Explanation:
Given that, the mass of the particle, [tex]m=2.4\times10^{-6} kg[/tex].
Charge of the particle, [tex]q=27\times10^{-6} kg[/tex].
Speed of the particle, [tex]v=27 m/s[/tex].
Magnetic field is, [tex]B=9.5 T[/tex].
The radius of the particle's circular path when the speed of the particle is perpendicular to the magnetic field is,
[tex]r=\frac{mv}{qB}[/tex].
By putting all the values in the above equation.
[tex]r=\frac{(2.4\times10^{-6}kg)(27m/s) }{(27\times10^{-6}C)(9.5T) }[/tex].
Therefore by further solving,
[tex]r=0.253m[/tex]
Therefore the radius of the particle in a circular path is 0.253 m
