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Two parallel wires of equal lengths are separated by a distance of $3 \mathrm{~m}$ from each
other. The currents flowing through first and second wire is $3 \mathrm{~A}$ and $4.5 \mathrm{~A}$
respectively in opposite directions. The resultant magnetic field at mid-point of both the wires is $\left(\mu_{0}=\right.$ permeability of free space)
Options:
other. The currents flowing through first and second wire is $3 \mathrm{~A}$ and $4.5 \mathrm{~A}$
respectively in opposite directions. The resultant magnetic field at mid-point of both the wires is $\left(\mu_{0}=\right.$ permeability of free space)
Solution:
2499 Upvotes
Verified Answer
The correct answer is:
$\frac{5 \mu_{0}}{2 \pi}$
The currents are in opposite direction and hence their fields are in same direction at the mid point which is at a distance of $1.5 \mathrm{~m}$ from each wire.
$\begin{aligned}
\mathrm{B}_{1} &=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1}}{\mathrm{r}}=\frac{\mu_{0}}{2 \pi} \cdot \frac{3}{1.5}=\frac{2 \mu_{0}}{2 \pi} \\
\mathrm{B}_{2} &=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{2}}{\mathrm{r}}=\frac{\mu_{0}}{2 \pi} \cdot \frac{4.5}{1.5}=\frac{3 \mu_{0}}{2 \pi} \\
\therefore \mathrm{B} &=\mathrm{B}_{1}+\mathrm{B}_{2}=\frac{5 \mu_{0}}{2 \pi}
\end{aligned}$
$\begin{aligned}
\mathrm{B}_{1} &=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1}}{\mathrm{r}}=\frac{\mu_{0}}{2 \pi} \cdot \frac{3}{1.5}=\frac{2 \mu_{0}}{2 \pi} \\
\mathrm{B}_{2} &=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{2}}{\mathrm{r}}=\frac{\mu_{0}}{2 \pi} \cdot \frac{4.5}{1.5}=\frac{3 \mu_{0}}{2 \pi} \\
\therefore \mathrm{B} &=\mathrm{B}_{1}+\mathrm{B}_{2}=\frac{5 \mu_{0}}{2 \pi}
\end{aligned}$
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