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This question is from mixture of modern physics and magnetism. Ans is10(2)^1/2

This question is from mixture of modern physics and magnetism. Ans is10(2)^1/2

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Grade:12th pass

1 Answers

Khimraj
3007 Points
3 years ago
radius of orbit in magnetic field is given by
r = mv/qB
Let potential difference is V
kinetic energy E = (½)mv2 = qV
so mv  = \sqrt{2mE} = \sqrt{2mqV}
So r = (1/B)\sqrt{2mV/q} \propto \sqrt{m/q}
m/q for  proton = 1/e
and for \alpha particle = 4/2e = 2/e
so radius for \alpha particle = \sqrt{2}times pf proton = 10\sqrt{2} cm.
Hope it clears. If you like answer then please aapprove it.

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