MHT CET · Physics · Magnetic Effects of Current
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)
- A \(\frac{3 \mu_{0}}{2 \pi}\)
- B \(\frac{7 \mu_{0}}{2 \pi}\)
- C \(\frac{\mu_{0}}{2 \pi}\)
- D \(\frac{5 \mu_{0}}{2 \pi}\)
Answer & Solution
Correct Answer
(D) \(\frac{5 \mu_{0}}{2 \pi}\)
Step-by-step Solution
Detailed explanation
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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