Answer:
[tex]1.5048\times 10^{-23}\ Am^2[/tex]
Explanation:
q = Charge of proton = [tex]1.6\times 10^{-19}\ C[/tex]
r = Radius of circle = [tex]5.7\times 10^{-11}\ m[/tex]
v = Velocity of proton = [tex]3.3\times 10^6\ m/s[/tex]
Magnetic moment is given by
[tex]M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2[/tex]
The magnetic moment associated with this motion is [tex]1.5048\times 10^{-23}\ Am^2[/tex]
