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An electron $$($$mass $$=9.1\times 10^{-31}$$; charge $$=-1.6\times 10^{-19}\mathrm{C})$$ experiences no deflection if subjected to an electric field of $$3.2\times 10^{5}\mathrm{V}/\mathrm{m}$$ and a magnetic field of $$2.0\times 10^{-3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$. Both the fields are normal to the path of electron and to each other. Ifthe electric field is removed, then the electron will revolve in an orbit of radius :


A
45m
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B
4.5m
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C
0.45m
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D
0.045m
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Solution

The correct option is C $$0.45m$$
$$V=\dfrac{E}{B} $$ For no deflection to occur
$$V=\dfrac{3.2\times 10^5}{2\times 10^{-3}}=1.6\times 10^8m/s$$
$$R=\dfrac{mv}{qB}$$
$$R=\dfrac{9.1\times10^{-31}\times1.6\times 10^8}{1.6\times10^{-19}\times 2\times10^{-3}}$$
$$R=0.45m$$

Physics

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