A charge of $1 \times 10^{- 2} C$ moves at 15 revolutions per second in a circle of radius $40 c m$ . The current linked with the circuit is

0.3 A

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0.15 A

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0.6 A

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0.03 A

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Solution

The correct option is B $0.15 A$
We know that,

$I = \frac{Q}{t}$
Where,
$I : E l e c t r i c C u r r e n t$
$Q : T o t a l C h a r g e$
$t : T i m e$
The charge in the circuit takes 1 s to complete 15 revolutions.

So for 1 revolution, it will take $1 / 15 s$ .

$I$ = $\frac{1 \times 10^{- 2}}{\frac{1}{15}}$ = $0.15 A$

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