### Magnetic Effects Of Current Question 8

#### Question 8 - 25 January - Shift 2

Two long parallel wires carrying currents $8 A$ and $15 A$ in opposite directions are placed at a distance of $7 cm$ from each other. A point $P$ is at equidistant from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnitude of magnetic field at $P$ is $\times 10^{-6} T$. (Given : $\sqrt{2}=1.4$ )

## Show Answer

#### Answer: (68)

#### Solution:

#### Formula: Magnetic field due to a straight wire

Magnetic fields due to both wires will be perpendicular to each other.

$B_1=\frac{\mu_0 i_1}{2 \pi d} \quad B_2=\frac{\mu_0 i_2}{2 \pi d}$

$B _{\text{net }}=\sqrt{B_1^{2}+B_2^{2}} \Rightarrow \frac{\mu_0}{2 \pi d} \sqrt{i_1^{2}+i_2^{2}}$

$\Rightarrow \frac{4 \pi \times 10^{-7}}{2 \pi \times(7 / \sqrt{2}) \times 10^{-2}} \times \sqrt{8^{2}+15^{2}}(d=\frac{7}{\sqrt{2}} cm)$

$\Rightarrow 68 \times 10^{-6} T$