Two long parallel straight metal wires A and B carrying currents $12 \mathrm{~A}$ and $36 \mathrm{~A}$…
- $90 \mathrm{~cm}$
- $7.5 \mathrm{~cm}$
- $28 \mathrm{~cm}$
- $12.5 \mathrm{~cm}$
Solution

Suppose at $\mathrm{P}, \mathrm{B}=0$ So, $\mathrm{B}=0$ $\begin{aligned} & \Rightarrow \frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}+\frac{\mu_0 \mathrm{i}_2}{2 \pi(0.5-\mathrm{r})}=0 \\ & \Rightarrow \frac{\mathrm{i}_1}{\mathrm{r}}+\frac{\mathrm{i}_2}{0.5-\mathrm{r}}=0 \\ & \Rightarrow \frac{12}{\mathrm{r}}+\frac{36}{0.5-\mathrm{r}} \\ & \Rightarrow 0.5-\mathrm{r}=3 \mathrm{r} \\ & \Rightarrow 0.5=4 \mathrm{r} \\ & =\mathrm{r}=\frac{1}{8} \mathrm{~m}=12.5 \mathrm{~cm} \end{aligned}$
Asked in: MHT CET Full Test 5
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