Answer:
r = 0.0548 m
Explanation:
Given that,
Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.
We need to find the radius of their circular path. The formula for the radius of path is given by :
[tex]r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}[/tex]
m is mass of Singly charged uranium-238 ion, [tex]m=3.95\times 10^{-25}\ kg[/tex]
q is charge
So,
[tex]r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m[/tex]
So, the radius of their circular path is equal to 0.0548 m.
