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)

Solution :

Magnetic field due to a long straight current carrying wire is :

$$B = \frac{\mu_0 I}{2\pi r}$$

Given :

Current in first wire,

$$I_1 = 8\text{ A}$$

Current in second wire,

$$I_2 = 15\text{ A}$$

Distance between wires,

$$d = 7\text{ cm}$$

Since point P is equidistant from both wires and the lines joining P to the wires are perpendicular to each other, triangle formed is right angled isosceles.

Hence, distance of P from each wire :

$$r = \frac{7}{\sqrt{2}}\text{ cm}$$

$$= \frac{7}{1.4}\text{ cm}$$

$$= 5\text{ cm} = 0.05\text{ m}$$

Magnetic field due to first wire :

$$B_1 = \frac{\mu_0 I_1}{2\pi r}$$

$$= \frac{4\pi \times 10^{-7} \times 8}{2\pi \times 0.05}$$

$$= 32 \times 10^{-6}\text{ T}$$

Magnetic field due to second wire :

$$B_2 = \frac{\mu_0 I_2}{2\pi r}$$

$$= \frac{4\pi \times 10^{-7} \times 15}{2\pi \times 0.05}$$

$$= 60 \times 10^{-6}\text{ T}$$

Since the magnetic fields are perpendicular,

$$B = \sqrt{B_1^2 + B_2^2}$$

$$= \sqrt{32^2 + 60^2}\times 10^{-6}$$

$$= \sqrt{4624}\times 10^{-6}$$

$$= 68 \times 10^{-6}\text{ T}$$

Final Answer :

$$68$$