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Question: If an electron is revolving in a circular orbit of radius \(0.5\,{A^ \circ }\) with a velocity of \(...

If an electron is revolving in a circular orbit of radius 0.5 A∘0.5\,{A^ \circ } with a velocity of 2.2×106 m/s.2.2 \times {10^6}\,m/s. The magnetic dipole moment of the revolving electron is
A. 8.8×10−24 Am8.8 \times {10^{ - 24}}\,Am
B. 8.8×10−23 Am8.8 \times {10^{ - 23}}\,Am
C. 8.8×10−22 Am8.8 \times {10^{ - 22}}\,Am
D. 8.8×10−21 Am8.8 \times {10^{ - 21}}\,Am

Explanation

Solution

Hint In the question, the circular orbit of radius and the velocity of the electron is given. By substituting the values in the equation of the magnetic dipole moment, we get the value of the magnetic dipole moment of the revolving electron.
Formula used
The expression for finding the magnetic dipole moment is
μ=i.A\mu = i.A
Where
iibe the tiny current travelling around the edge and AAbe the cross sectional area of the flowing of current.

Complete step by step solution
Given that μ=0.5 A∘ and v=2.2×106 ms−1\mu = 0.5\,{A^ \circ }\,{\text{and }}v = 2.2 \times {10^6}\,m{s^{ - 1}}
Magnetic dipole moment μ=i.A............(1)\mu = i.A............\left( 1 \right)
The charge of electron is constant e=1.6×10−19e = 1.6 \times {10^{ - 19}}
Current travelling around the edge i=efi = ef
Where,
eebe the charge of the electron ,ffbe the frequency and the wwbe the wavelength.
Cross sectional area A=Ar2A = A{r^2}
Substitute the parameters in the equation (1)\left( 1 \right)
μ=ef.Ar2\mu = ef.A{r^2}
μ=ew2a.Ar2\mu = \dfrac{{ew}}{{2a}}.A{r^2}
Simplify the above equation, we get
μ=evr22r\mu = \dfrac{{ev{r^2}}}{{2r}}
Simplify the above equation, we get
μ=evr2\mu = \dfrac{{evr}}{2}
Substitute the known values in the above equation, we get
μ=1.6×10−19×2.2×106×0.5×10−102\mu = \dfrac{{1.6 \times {{10}^{ - 19}} \times 2.2 \times {{10}^6} \times 0.5 \times {{10}^{ - 10}}}}{2}
Simplify the above equation we get,
μ=8.8×10−24 Am\mu = 8.8 \times {10^{ - 24}}\,Am
Therefore, the magnetic dipole moment of the electron is 8.8×10−24 Am8.8 \times {10^{ - 24}}\,Am.

Hence from the above options, option A is correct.

Note In the question, we have to find the magnetic dipole moment. But in this case, suppose we have to find the cross sectional area we have to apply the formula for the cross sectional area based on the shape of the electron. The value of the magnetic dipole moment varies based on the shape.