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Question

In Bohr's model of hydrogen atom, the electron circulates round the nucleus in a path of radius $$5\times 10^{-11}$$m at a frequency of $$6.8\times 10^{15}$$ revolutions per second. The value of magnetic induction at the centre of the orbit is :


A
12.27T
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B
10.8T
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C
13.2T
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D
13.6T
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Solution

The correct option is A $$13.6T$$

 $$\left | B  \right |$$  =$$\dfrac{\mu _{0} i}{2r}$$

I=$$\dfrac{\mathrm{d} q}{\mathrm{d} t}$$

= $$qf$$

= $$1.6\times  10^{-19}\times  6.8\times 10^{15}$$

=$$10.88\times  10^{-4}$$

$$\therefore \left | B \right | = \dfrac{4\pi\times 10^{-7}\times 10.88\times  10^{-4}}{2 \times 5 \times 10^{-11}}$$

=$$13.6T$$

Physics

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