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NEET · Physics · STD 12 - 4. Moving charges and magnetism
A straight conductor carrying current \(i\) splits into two parts as shown in the figure. The radius of the circular loop is \(R\). The total magnetic field at the centre \(P\) of the loop is

- A \(0\)
- B \(\frac {3 \mu_{0} i} {32 R}\), outward
- C \(\frac {3 \mu_{0} i} {32 \mathrm{R}},\) Inward
- D \(\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}},\) inward
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\ell_{1}}{\ell_{2}}= \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{\theta_{1}}{\theta_{2}} \Rightarrow \theta_{1} \mathrm{I}_{1}=\theta_{2} \mathrm{I}_{2}\) \(\left.\begin{array}{l}{\mathrm{B}_{1}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{1} \theta_{1}}{\mathrm{r}}} \\ {\mathrm{B}_{2}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{1} \theta_{1}}{\mathrm{r}}}\end{array}\right\} \Rightarrow \mathrm{B}_{1}=\mathrm{B}_{2}\)


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