An electron in a circular orbit of radius 0.0 | vedclass

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(D) The magnetic moment $M$ of a current loop is given by $M = I 	imes A$. Here, the current $I$ due to the revolving electron is $I = qf = ef$, wher

Vedclass · Accepted Answer

$1.26 	imes 10^{-23} \,A-m^2$

Vedclass · Answer

(D) The magnetic moment $M$ of a current loop is given by $M = I 	imes A$. Here, the current $I$ due to the revolving electron is $I = qf = ef$, where $f$ is the frequency of revolution. The area $A$ of the circular orbit is $A = \pi r^2$. Substituting these into the formula, we get $M = (ef)(\pi r^2)$. Given: $r = 0.05 \,nm = 0.05 	imes 10^{-9} \,m = 5 	imes 10^{-11} \,m$, $f = 10^{16} \,Hz$, and $e = 1.6 	imes 10^{-19} \,C$. $M = (1.6 	imes 10^{-19} \,C) 	imes (10^{16} \,s^{-1}) 	imes (3.14) 	imes (5 	imes 10^{-11} \,m)^2$. $M = 1.6 	imes 10^{-3} 	imes 3.14 	imes 25 	imes 10^{-22}$. $M = 1.6 	imes 3.14 	imes 25 	imes 10^{-25}$. $M = 125.6 	imes 10^{-25} = 1.256 	imes 10^{-23} \,A-m^2 \approx 1.26 	imes 10^{-23} \,A-m^2$.

An electron in a circular orbit of radius $0.05 \,nm$ performs $10^{16}$ revolutions per second. What is the magnetic moment due to the rotation of the electron? $(e = 1.6 \times 10^{-19} \,C)$

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