Answer:
Radius of the path will be [tex]20.136\times 10^{-4}m[/tex]
Step-by-step explanation:
We have given mass of the singly charged positive ion [tex]m=3.20\times 10^{-26}kg[/tex]
Potential difference V = 858 volt
Magnetic force on the charge particle is given by [tex]F=qvB[/tex], here q is charge v is velocity and B is magnetic field
And centripetal force is given by [tex]F=\frac{mv^2}{r}[/tex]
These forces will be equal
So [tex]\frac{mv^2}{r}=qvB[/tex]
[tex]r=\frac{mv}{qB}[/tex]
We know that [tex]V=\frac{W}{q}=\frac{\frac{1}{2}mv^2}{q}[/tex]
[tex]v=\sqrt{\frac{2Vq}{m}}=\sqrt{\frac{2\times 858\times 1.6\times 10^{-19}}{3.2\times 10^{-26}}}=9.2628\times 10^{3}m/sec[/tex]
Now radius [tex]r=\frac{mv}{qB}=\frac{3.2\times 10^{-26}\times 9.2628\times 10^{3}}{1.6\times 10^{-19}\times 0.920}=20.136\times 10^{-4}m[/tex]
