Two wires carrying currents of 5 A and 2 A | vedclass

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(C) According to Ampere's circuital law,the line integral of the magnetic field $\overrightarrow{B}$ around any closed loop is equal to $\mu_0$ times

Vedclass · Accepted Answer

$3 \mu_0$

Vedclass · Answer

(C) According to Ampere's circuital law,the line integral of the magnetic field $\overrightarrow{B}$ around any closed loop is equal to $\mu_0$ times the net current $I_{	ext{enclosed}}$ passing through the loop.$\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I_{	ext{enclosed}}$In the given figure,two wires carrying currents of $5 \ A$ and $2 \ A$ are inside the loop. Since these currents are in opposite directions,the net current enclosed by the loop is $I_{	ext{enclosed}} = 5 \ A - 2 \ A = 3 \ A$.The wire carrying $3 \ A$ current is situated outside the loop,so it does not contribute to the net enclosed current.Therefore,the value of the line integral is $\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 	imes 3 \ A = 3 \mu_0$.

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