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An electron in a hydrogen atom revolves around its nucleus with a speed of $$6.76 \times 10^6$$ m s$$^{-1}$$ in an orbit of radius $$0.52$$ A. The magnetic field produced at the nucleus of the hydrogen atom is _____ T.
Correct Answer: 40
$$I = \frac{q}{T} = \frac{ev}{2\pi r}$$
$$B = \frac{\mu_0 I}{2r}$$
$$B = \frac{\mu_0 e v}{4\pi r^2}$$
$$B = \frac{(4\pi \times 10^{-7}) \cdot (1.6 \times 10^{-19}) \cdot (6.76 \times 10^6)}{4\pi \cdot (0.52 \times 10^{-10})^2}$$
$$B = 40 \text{ T}$$
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