Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each

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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…

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Given: Two long parallel wires carry currents $i_1 = 8 \, \text{A}$ and $i_2 = 15 \, \text{A}$ in opposite directions. The separation between the wires is $7 \, \text{cm}$ .

Find: The magnitude of magnetic field at point P in the form _____ x $10^{-6} \, \text{T}$ .

Right triangle arrangement of two long wires and point P, with currents 8 A and 15 A, wire separation 7 cm, and equal perpendicular distances d from P to each wire.

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

For each long straight wire,

B_1 = \frac{\mu_0 i_1}{2\pi d}, \qquad B_2 = \frac{\mu_0 i_2}{2\pi d}

Since the two magnetic fields are perpendicular,

B_{\text{net}} = \sqrt{B_1^2 + B_2^2}

Therefore,

B_{\text{net}} = \frac{\mu_0}{2\pi d}\sqrt{i_1^2 + i_2^2}

From the geometry, the distances from P to both wires are equal and perpendicular, so

d = \frac{7}{\sqrt{2}} \, \text{cm} = \left(\frac{7}{\sqrt{2}}\right) \times 10^{-2} \, \text{m}

Now substitute:

B_{\text{net}} = \frac{4\pi \times 10^{-7}}{2\pi \times \left(\frac{7}{\sqrt{2}}\right) \times 10^{-2}} \times \sqrt{8^2 + 15^2}

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

Therefore, the required value is $68$ .

Given: Point P is equidistant from the two wires, and the lines joining P to the wires are perpendicular.

Find: Net magnetic field magnitude at P.

The figure forms a right triangle with equal legs $d$ and hypotenuse $7 \, \text{cm}$ . Hence,

\sqrt{d^2 + d^2} = 7 \, \text{cm}

\sqrt{2d^2} = 7 \, \text{cm}

d = \frac{7}{\sqrt{2}} \, \text{cm}

Now, magnetic field due to a long straight wire is

B = \frac{\mu_0 I}{2\pi d}

So,

B_1 = \frac{\mu_0 \times 8}{2\pi d}, \qquad B_2 = \frac{\mu_0 \times 15}{2\pi d}

Because the directions of these two fields at P are perpendicular, use Pythagoras for vector addition:

B_{\text{net}} = \sqrt{B_1^2 + B_2^2}

B_{\text{net}} = \frac{\mu_0}{2\pi d} \sqrt{8^2 + 15^2}

Using

\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

and

d = \left(\frac{7}{\sqrt{2}}\right) \times 10^{-2} \, \text{m}

we get the final magnetic field as

68 \times 10^{-6} \, \text{T}

Therefore, the required numerical answer is $68$ .

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