Question

If the magnitude of the magnetic field of induction at point $S$ is given by $x\times {10}^{-7}T$. Find x.

Answer 1

Since, ${AS}^2+{BS}^2={AB}^2$
$\therefore -ASB={90}^o$

At point $S$:

$B_1=$ Magnetic field due to $i_A$
$=\frac{{\mu }_0}{2\pi }\frac{i_A}{1.6}=\frac{(2\times {10}^{-7})\left(9.6\right)}{1.6}=12\times {10}^{-7}T$

$B_2=$ Magnetic field due to $i_B$
$=\frac{{\mu }_0}{2\pi }\frac{i_B}{1.2}=\frac{(2\times {10}^{-7})\left(3\right)}{1.2}=5\times {10}^{-7}T$

Since $B_1$ and $B_2$ are mutually perpendicular, net magnetic field at$S$ would be:
$B=\sqrt{{B_1}^2+{B_2}^2}=\sqrt{{(12\times {10}^{-7})}^2+{(5\times {10}^{-7})}^2}$
$=13\times {10}^{-7}T$