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 \(1^{\text {st }}\) and \(2^{\text {nd }}\) wire is \(3 \mathrm{~A}\) and \(4.5 \mathrm{~A}\) respectively in opposite directions. The resultant magnetic field at mid point between the wires \(\left(\mu_0=\right.\) permeability of free space)

A \(\frac{\mu_0}{2 \pi}\)
B \(\frac{3 \mu_0}{2 \pi}\)
C \(\frac{7 \mu_0}{2 \pi}\)
D \(\frac{5 \mu_0}{2 \pi}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{5 \mu_0}{2 \pi}\)

Step-by-step Solution

Detailed explanation

Using Biot Savart law,
\(\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}\)
\(\therefore \quad\) Magnetic field due to first wire:
\(\mathrm{B}_1=\frac{\mu_0 \mathrm{I}_1}{2 \pi \mathrm{r}}=\frac{\mu_0 \times 3}{2 \pi \times 1.5}=\frac{2 \mu_0}{2 \pi}\)
\(\therefore \quad\) Magnetic field due to second wire:
\(\mathrm{B}_2=\frac{\mu_{\mathrm{o}} \mathrm{I}_2}{2 \pi \mathrm{r}}=\frac{\mu_{\mathrm{o}} \times 4.5}{2 \pi \times 1.5}=\frac{3 \mu_0}{2 \pi}\)
\(\therefore \quad\) Net field,
\(\mathrm{B}=\mathrm{B}_1+\mathrm{B}_2=\frac{2 \mu_0}{2 \pi}+\frac{3 \mu_0}{2 \pi}=\frac{5 \mu_0}{2 \pi}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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