A helium nucleus makes a full rolation in a circle of radius 0.8 metre in two seconds. The value of the magnetic field B at the centre of the circle will be?
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The current deu to revolution of the HE Nucleus is I = qw/2π = 2 x 1.6 x 10^-19 / 2 =
1.6 x 10^-19 A
Magnetic field at the center B = βoI / 2r =
βo x 1.6 x 10^-19 = 10^-19βo
So 10^-19βo is your answer.
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