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(D) The magnetic field $B$ at the center of a circular current loop is given by the formula: $B = \frac{\mu_0 i}{2r}$.Given values are: $B = 12.56 \,

Vedclass · Accepted Answer

$1.04 	imes 10^{-3} \, A$

Vedclass · Answer

(D) The magnetic field $B$ at the center of a circular current loop is given by the formula: $B = \frac{\mu_0 i}{2r}$.Given values are: $B = 12.56 \, T$,$r = 5.2 	imes 10^{-11} \, m$,and $\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$.Substituting these values into the formula:$12.56 = \frac{(4\pi 	imes 10^{-7}) 	imes i}{2 	imes 5.2 	imes 10^{-11}}$$12.56 = \frac{2\pi 	imes 10^{-7} 	imes i}{5.2 	imes 10^{-11}}$Using $\pi \approx 3.14$,we have $2\pi \approx 6.28$.$12.56 = \frac{6.28 	imes 10^{-7} 	imes i}{5.2 	imes 10^{-11}}$$i = \frac{12.56 	imes 5.2 	imes 10^{-11}}{6.28 	imes 10^{-7}}$$i = 2 	imes 5.2 	imes 10^{-4} \, A$$i = 10.4 	imes 10^{-4} \, A = 1.04 	imes 10^{-3} \, A$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

In a hydrogen atom,an electron moves in a circular orbit of radius $5.2 \times 10^{-11} \, m$ and produces a magnetic induction of $12.56 \, T$ at its nucleus. The current produced by the motion of the electron will be (Given $\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m$)

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