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

Let's determine the magnetic field at point P due to two parallel wires carrying currents in opposite directions. Given that the currents are $I_1 = 8 \text{ A}$ and $I_2 = 15 \text{ A}$, and the distance between the wires is $d = 0.07 \text{ m}$. Point P is equidistant from both wires, with the lines joining P to each wire perpendicular to one another.
By symmetry and geometry, each distance from P to a wire can be defined as $r = \frac{d}{\sqrt{2}} = \frac{0.07}{1.4} = 0.05 \text{ m}$.
The magnetic field due to a long straight wire at a distance r is given by \(B = \frac{\mu_0 I}{2\pi r}\). Here, \(\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}\).
Calculating the magnetic field due to the first wire: 
\(B_1 = \frac{4\pi \times 10^{-7} \times 8}{2\pi \times 0.05} = \frac{32 \times 10^{-7}}{0.1} = 32 \times 10^{-6} \text{ T}\).
Calculating the magnetic field due to the second wire: 
\(B_2 = \frac{4\pi \times 10^{-7} \times 15}{2\pi \times 0.05} = \frac{60 \times 10^{-7}}{0.1} = 60 \times 10^{-6} \text{ T}\).
Since the currents are in opposite directions, the magnetic fields add up perpendicularly. Thus, the resultant magnetic field at P is calculated using Pythagorean theorem:
\(B = \sqrt{B_1^2 + B_2^2} = \sqrt{(32 \times 10^{-6})^2 + (60 \times 10^{-6})^2}\). 
\(B = \sqrt{1024 \times 10^{-12} + 3600 \times 10^{-12}} = \sqrt{4624 \times 10^{-12}} = 68 \times 10^{-6} \text{ T}\).
Thus, the magnitude of the magnetic field at point P is \(68 \times 10^{-6} \text{ T}\), which is within the specified range.