If $α$ -particle $(_{2} {He}^{4})$ is revolving in a circular orbit of radius 3.14 $\overset{0}{A}$ with speed of 8 x 10⁶ m/s. Then the equivalent current is

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$12.8 \times 10^{- 4} A$

$6.4 \times 10^{- 12} A$

$12.8 \times 10^{- 16} A$

$6.4 \times 10^{- 9} A$

answer is B.

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Detailed Solution

$\begin{array}{l} i = \frac{q}{t} i = \frac{qV}{2 π r} = \frac{(2) 1.6 \times 10^{- 19} \times 8 \times 10^{6}}{2 \times 3.14 \times 10^{- 10} \times 3.14} \\ = 12.8 \times 10^{- 4} A \end{array}$

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