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  • In a Helium atom, the electron is making 12.6 * 1015 revolutions per second in a circular orbit of radius 0.49 Å. The field of induction produced at the positi

In a Helium atom, the electron is making 12.6 * 1015 revolutions per second in a circular orbit of radius 0.49 Å. The field of induction produced at the position of the nucleus in Tesla is:

Option: 1

6\piT


Option: 2

7\piT


Option: 3

8\piT


Option: 4

9\piT


Answers (1)

best_answer

W = 2\pi r = 2 \pi * 12.6 * 10^{15}

B = \frac{\mu_{0}i}{2R} = \frac{qW}{2 \pi}

= \frac{4\pi * 10^{-7} * 1.6 * 10^{-19} * 12.6 * 10^{15}}{2 * 0.49 * 10^{-10}}

= \frac{4* 1.6 * 12.6}{2 * 0.49} * \pi * 10^{-7-19+15+10}

= \frac{80.64}{0.98} * \pi * 10^{25 - 26}

= \frac{80.64}{0.98} * \pi * 10^{-1}

= \frac{80.64}{9.8} * \pi = 8.22 \pi \approx 8\pi T

Posted by

Suraj Bhandari

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