Two wires with currents 2 A$\ $and $1$ A are enclosed in a circular loop. Another wire with current 3 A is situated outside the loop as shown. The $\oint_{}^{}{\overrightarrow{B}.d\overrightarrow{l}}$ around the loop is [image:4_image24.emf]

Question

Dr. Abubacker · Accepted Answer

(a) According to Ampere’s circuital law
$\oint_{}^{}{\overrightarrow{B}.d\overrightarrow{l}} = \mu_{0}I_{enclosed} = \mu_{0}\lbrack 2A - 1A\rbrack = \mu_{0}$
Since currents $2A$ and $1A$ are in the opposite direction

Q. Two wires with currents 2 A$\ $and $1$ A are enclosed in a circular loop. Another wire with current 3 A is situated outside the loop as shown. The $\oint_{}^{}{\overrightarrow{B}.d\overrightarrow{l}}$ around the loop is

Explanation

Reviewed by Dr. Abubacker

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