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Ampere's Circuital Law

A closed curv...

Question

A closed curve encircles several conductors.The line integral $\int \to B \cdot d \to l$ around this curve is $3.83 \times 10^{- 7} T - m$ . What is the net current in the conductors?

A

0.1 A

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B

0.2 A

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C

0.3 A

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D

0.4 A

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Solution

The correct option is C $0.3 A$
From Ampere's law, for a closed loop carrying current
$\int \to B . d \to l = μ_{0} i$
$i = \frac{\int \to B . d \to l}{μ_{0}}$
$= \frac{3.83 \times 10^{- 7}}{μ_{0}}$
$= 0.3 A$

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