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Q.

An electron is revolving around a proton in a circular orbit of diameter 1 Å. If it produces a magnetic field of 14 weber/m2 at the proton, then its angular velocity will be about

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a

4x 1016 rad,/sec

b

1016 rad/sec

c

4x 1015 rad,sec

d

1015 rad/sec

answer is A.

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Detailed Solution

B=μ04π×2πir=μ04π×2πet×1r=μ04π×2π×e(2πr/v)×1r=μ04π×2π×ev2πr×1r=μ04π×eωr∴ ω=4πμ0×rBe=107×10−10×142×1⋅6×10−19=4×1016rad/sec
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