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A singly charged ion of 7Li (an isotope of lithium) has a mass of 1.16×10−26 kg . It is accelerated through a potential difference of 290 V and then enters a magnetic field with magnitude 0.710 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

Respuesta :

Answer:

0.00645 m

Explanation:

m = Mass of lithium = [tex]1.16\times 10^{-26}\ kg[/tex]

V = Voltage = 290 V

r = Radius of path

B = Magnetic field = 0.710 T

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

v = Velocity of mass

The kinetic energy in moving a charge and the work done by the charge will balance each other

[tex]\frac{1}{2}mv^2=qV\\\Rightarrow v=\sqrt{\frac{2qV}{m}}\\\Rightarrow v=\sqrt{\frac{1\times 1.6\times 10^{-19}\times 290}{1.16\times 10^{-26}}}\\\Rightarrow v=63245.5532\ m/s[/tex]

The velocity by which the charge is moving is 63245.5532 m/s

The force on moving a charge and the centripetal force will balance each other

[tex]F=F_c\\\Rightarrow Bqv=m\frac{v^2}{r}\\\Rightarrow r=\frac{mv}{Bq}\\\Rightarrow r=\frac{1.16\times 10^{-26}\times 63245.5532}{0.71\times 1.6\times 10^{-19}}\\\Rightarrow r=0.00645\ m[/tex]

The radius of the magnetic field is 0.00645 m