Answer:
The radius of curvature of the ion's orbit is 0.59 meters
Explanation:
Given that,
Mass of the 24 Mg ion, [tex]m=3.983\times 10^{-26}\ kg[/tex]
Potential difference, V = 3 kV
Magnetic field, B = 526 G
Charge on single ionized ion, [tex]q=1.6\times 10^{-19}\ C[/tex]
The radius of the the path traveled by the charge is circular. Its radius is given by :
[tex]r=\dfrac{mv}{Bq}[/tex]
v is speed of particle.
v can be calculated using conservation of energy as :
[tex]\dfrac{1}{2}mv^2=qV\\\\v=\sqrt{\dfrac{2qV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^3}{3.983 \times 10^{-26}}} \\\\v=1.26\times 10^5\ m/s[/tex]
Radius,
[tex]r=\dfrac{3.983 \times 10^{-26}\times 1.26\times 10^5}{0.0526\times 1.6\times 10^{-19}}\\\\r=0.59\ m[/tex]
So, the radius of curvature of the ion's orbit is 0.59 meters.
