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A current ' $I^{\prime}$ is flowing along an infinite, straight wire, in the positive $Z$ -direction and the same current is flowing along a similar parallel wire $5 \mathrm{m}$ apart, in the negative Z-direction. A point $P$ is at a perpendicular distance $3 \mathrm{m}$ from the first wire and $4 \mathrm{m}$ from the second. What will be magnitude of the magnetic field B of $P ?$
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Verified Answer
The correct answer is:
$\frac{5}{24}\left(\mu_{0} I\right)$
According to the question.
Magnetic field due to first wire $(A)$ is $B_{1}=\frac{\mu_{0} \times I}{2 \pi \times 3}$
Magnetic field due to second wire $(B)$ is $B_{2}=\frac{\mu_{0} \times 1}{2 \pi \times 4}$
Net magnitude of magnetic field $B=\sqrt{B_{1}^{2}+B_{2}^{2}}$
$$
B=\frac{\mu_{0} I}{2 \pi} \sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{\mu_{0} I \times 5}{2 \pi \times 3 \times 4}
$$
Magnetic field $B=\frac{5}{24} \times \frac{\mu_{0} I}{\pi}$
Magnetic field due to first wire $(A)$ is $B_{1}=\frac{\mu_{0} \times I}{2 \pi \times 3}$
Magnetic field due to second wire $(B)$ is $B_{2}=\frac{\mu_{0} \times 1}{2 \pi \times 4}$
Net magnitude of magnetic field $B=\sqrt{B_{1}^{2}+B_{2}^{2}}$
$$
B=\frac{\mu_{0} I}{2 \pi} \sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{\mu_{0} I \times 5}{2 \pi \times 3 \times 4}
$$
Magnetic field $B=\frac{5}{24} \times \frac{\mu_{0} I}{\pi}$
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