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(D) The magnetic field $B$ at the center of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.The current $I$ produced by an e

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$40$

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(D) The magnetic field $B$ at the center of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.The current $I$ produced by an electron revolving with speed $v$ in an orbit of radius $r$ is $I = \frac{ev}{2\pi r}$.Substituting $I$ into the magnetic field formula: $B = \frac{\mu_0}{2r} \left( \frac{ev}{2\pi r} ight) = \frac{\mu_0 ev}{4\pi r^2}$.Given values: $e = 1.6 	imes 10^{-19} \, C$,$v = 6.76 	imes 10^6 \, m/s$,$r = 0.52 	imes 10^{-10} \, m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.$B = 10^{-7} 	imes \frac{1.6 	imes 10^{-19} 	imes 6.76 	imes 10^6}{(0.52 	imes 10^{-10})^2}$.$B = 10^{-7} 	imes \frac{10.816 	imes 10^{-13}}{0.2704 	imes 10^{-20}} = 10^{-7} 	imes 40 	imes 10^7 = 40 \, T$.

An electron in a hydrogen atom revolves around its nucleus with a speed of $6.76 \times 10^6 \, m/s$ in an orbit of radius $0.52 \, \mathring{A}$. The magnetic field produced at the nucleus of the hydrogen atom is $...... \, T$.

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