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An electron is revolving round a proton, producing a magnetic field of `16 weber//m^(2)` in a circular orbit of radius `1 Å. Its angular velocity will beA. `10^(17)rad//sec`B. `1//2pixx10^(12)rad//sec`C. `2pixx10^(12)rad//sec`D. `4pixx10^(12)rad//sec` |
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Answer» Correct Answer - A Magnetic field due to revoluton of electron `B=(mu_(0))/(4pi).(2pii)/(r)=(mu_(0))/(4pi).(2pi.((eomega)/(2pi)))/(r)=10^(-7)xx(eomega)/(r)` ` implies16=10^(-7)xx(1.6xx10^(-19)omega)/(1xx10^(-10))impliesomega=10^(17)rad//sec` |
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