Question

Two identical wires $R$ and $S$ lie parallel in the horizontal plane, their axis being $0.10m$ apart. A current of $30A$ in $S$ Neglecting the effect of the earth's magnetic flux density at a point $P$ in the plane of the wires. If $P$ is mid way between $R$ and $S[$permeability of free space ${\mu }_0=4\pi \times {10}^{-7}H{m}^{-1}]$

If total flux density is $1.6\times {10}^{-X}T$. Find X?

Answer 1

$\left(i\right)$ Midway between $R$ and $S$
Field due to R, $B_R=\frac{{\mu }_{0}I}{2\pi r}$
Field due to S, $B_S=\frac{{\mu }_{0}I}{2\pi r}$
$=\frac{4\pi \times {10}^{-7}\times 30}{2\pi \times 0.05}=12\times {10}^{-5}T$
$\because$ Direction of B due to two wires are opposite,
$B=B_R+B_S=0$
Hence, total flux density is zero.
$\left(ii\right)$ If $P$ is $0.05m$ from $R$ and $0.15m$ from $S$
$B_R=\frac{{\mu }_{0}I}{2\pi r}=\frac{4\times {10}^{-7}}{2\pi \times 0.05}$
$=4\times {10}^{-5}T$
$B_S=\frac{{\mu }_{0}I}{2\pi r}=\frac{4\pi \times {10}^{-7}\times 30}{2\pi \times 0.15}$
$\because B_R$ and $B_S$ are in the same direction
$\therefore$ Resultant field $=B_S+B_R=2\times {10}^{-7}(\frac{10}{0.05}+\frac{30}{0.15})=8\times {10}^{-4}T$
Hence, total flux density is $8\times {10}^{-4}Weber$