Question

If $\overrightarrow{B_1},\overrightarrow{B_2}$ and $\overrightarrow{B_3}$ are the magnetic field due to the wires carrying current $I_1,I_2$ and $I_3$, then in the expression of Ampere's circuital law, $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_{0}I$, $\overrightarrow{B}$ is

• A. $\overrightarrow{B}=\overrightarrow{B_1}-\overrightarrow{B_2}$
• B. $\overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$
• C. $\overrightarrow{B}=\overrightarrow{B_1}-\overrightarrow{B_2}+\overrightarrow{B_3}$
• D. $\overrightarrow{B}=\overrightarrow{B_3}$

Answer 1

The correct option is B $\overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$

In Ampere's circuital law,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_{enclosed}$

$I_{enclosed}=$ net current passing through the loop

$\overrightarrow{B}=$ resultant magnetic field due to all currents existing anywhere either inside or outside the loop

$\therefore \overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$

Hence, option $\left(b\right)$ is the correct answer.