Question

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 : $\sqrt{2}=1.4$ )

Answer 1

Correct Answer: 68

The correct answer is 68

Magnetic fields due to both wires will be perpendicular to each other.
$B_1=\frac{{\mu }_0{i}_1}{2\pi d}{B}_2=\frac{{\mu }_0{i}_2}{2\pi d}$
$B_{net}=\sqrt{B_1^2+B_2^2}\Rightarrow \frac{{\mu }_0}{2\pi d}\sqrt{i_1^2+i_2^2}$
$\Rightarrow \frac{4\pi \times 10^{-7}}{2\pi \times (7/\sqrt{2})\times 10^{-2}}\times \sqrt{8^2+15^2}\left(d=\frac{7}{\sqrt{2}}cm\right)$
$\Rightarrow 68\times 10^{-6}T$

Answer 2

The magnetic field produced by each wire at point P can be calculated using the Biot-Savart Law: \[ B = \frac{\mu_0 I}{2\pi d} \] For the 8A current: \[ B_1 = \frac{\mu_0 \times 8}{2\pi \times 7 \times 10^{-2}} \] For the 15A current: \[ B_2 = \frac{\mu_0 \times 15}{2\pi \times 7 \times 10^{-2}} \] Since the currents are flowing in opposite directions and point P is equidistant from both, the net magnetic field \( B_{\text{net}} \) is the vector sum of \( B_1 \) and \( B_2 \): \[ B_{\text{net}} = \sqrt{B_1^2 + B_2^2} \Rightarrow \frac{\mu_0}{2\pi d} \sqrt{i_1^2 + i_2^2} \] \[ \Rightarrow \frac{4\pi \times 10^{-7}}{2\pi \times \left(\frac{7}{\sqrt{2}}\right) \times 10^{-2}} \times \sqrt{8^2 + 15^2} \quad \left( d = \frac{7}{\sqrt{2}} \, \text{cm} \right) \] Using the approximation \( \sqrt{2} = 1.4 \), we find that \( B_{\text{net}} = 68 \times 10^{-6} \, \text{T} \).