The radius of the first orbit of a hydrogen a | vedclass

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(D) Given: Radius $r = 0.5 \, \mathring{A} = 0.5 	imes 10^{-10} \, m$,Velocity $v = 2.2 	imes 10^6 \, m/s$,Charge $e = 1.6 	imes 10^{-19} \, C$.The

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$14.08$

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(D) Given: Radius $r = 0.5 \, \mathring{A} = 0.5 	imes 10^{-10} \, m$,Velocity $v = 2.2 	imes 10^6 \, m/s$,Charge $e = 1.6 	imes 10^{-19} \, C$.The magnetic field $B$ at the center (proton) due to a revolving electron is given by $B = \frac{\mu_0 I}{2r}$.Since $I = \frac{e}{T}$ and $T = \frac{2\pi r}{v}$,we have $I = \frac{ev}{2\pi r}$.Substituting $I$ into the formula for $B$: $B = \frac{\mu_0}{2r} \left( \frac{ev}{2\pi r} ight) = \frac{\mu_0 ev}{4\pi r^2}$.Using $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$:$B = \frac{10^{-7} 	imes (1.6 	imes 10^{-19}) 	imes (2.2 	imes 10^6)}{(0.5 	imes 10^{-10})^2}$.$B = \frac{10^{-7} 	imes 1.6 	imes 10^{-19} 	imes 2.2 	imes 10^6}{0.25 	imes 10^{-20}}$.$B = \frac{3.52 	imes 10^{-20}}{0.25 	imes 10^{-20}} = 14.08 \, T$.

The radius of the first orbit of a hydrogen atom is $0.5 \, \mathring{A}$ and the speed of the electron is $2.2 \times 10^6 \, m/s$. Find the magnetic induction at the proton due to the motion of the electron in $Tesla$.

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