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  • A helium nucleus makes a full rotation in a circle of radius 0.8 m in two seconds. The value of magnetic field B at the centre of the circle will be:Option: 1</s

A helium nucleus makes a full rotation in a circle of radius 0.8 m in two seconds. The value of magnetic field B at the centre of the circle will be:

Option: 1

10^{-19} / \mu_0


Option: 2

10^{-19} \mu_0


Option: 3

2 \times 10^{-19} \mu_0


Option: 4

2 \times 10^{-19} / \mu_0


Answers (1)

\mathrm{ q=2 e \\ }

\mathrm{ i=\frac{2 e \times 2}{2 \pi \times 0.8} \\ }

\mathrm{ B=\frac{\mu_0 l}{2 r} }.

Posted by

Kshitij

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