Question

Two infinitely long linear conductors are arranged perpendicular to each other and are in mutually perpendicular planes as shown in the figure. If $I_1=2$ $A$ along $y–axis$ and $I_2=3$ $A$ along $-vez–\text{axis}$ and $AP=AB=1cm$, the value of magnetic field strength $\overrightarrow{B}$ at $P$ is

• A. $(-3\times {10}^{-5}T)\hat{j}+(4\times {10}^{-5}T)\hat{k}$
• B. $(3\times {10}^{-5}T)\hat{j}+(4\times {10}^{-5}T)\hat{k}$
• C. $(4\times {10}^{-5}T)\hat{j}+(3\times {10}^{-5}T)\hat{k}$
• D. $(3\times {10}^{-5}T)\hat{j}+(-4\times {10}^{-5}T)\hat{k}$

Answer 1

The correct option is B $(3\times {10}^{-5}T)\hat{j}+(4\times {10}^{-5}T)\hat{k}$

$B$ due to $I_1$
Magnetic field due to $I_1$ at $P$ will be

$B_1=\frac{{\mu }_0{i}_1}{2\pi \left(r\right)}\hat{k}=\frac{{\mu }_0{i}_1}{2\pi \left(AP\right)}\hat{k}=\frac{2\times {10}^{-7}\times 2}{{10}^{-2}}\hat{k}$

(direction according to Maxwell’s right- hand thumb rule)
$B_1=4\times {10}^{-5}\hat{k}$

$B$ due to $I_2$

Magnetic field due to $I_2$ at $P$ will be

$B_2=\frac{{\mu }_0{i}_2}{2\pi \left(r\right)}\hat{j}=\frac{{\mu }_0{i}_2}{2\pi \left(PB\right)}\hat{j}=\frac{2\times {10}^{-7}\times 3}{2\times {10}^{-2}}\hat{j}$

(direction according to Maxwell’s right- hand thumb rule)
$B_2=3\times {10}^{-5}\hat{j}$

net magnetic field at $P$
${\overrightarrow{B}}_{\text{net}}={\overrightarrow{B}}_1+{\overrightarrow{B}}_2$

$=(3\times {10}^{-5}\hat{j}+4\times {10}^{-5}\hat{k})$

Hence, option (B) is correct.