A) \[\overrightarrow{dB}=\frac{{{\mu }_{0}}}{4\pi }\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r}\]
B) \[\overrightarrow{dB}=\frac{{{\mu }_{0}}}{4\pi }\frac{\overrightarrow{dl}\times \hat{r}}{{{r}^{3}}}\]
C) \[\overrightarrow{dB}=\frac{{{\mu }_{0}}}{4\pi }\frac{\overrightarrow{dl}\times \vec{r}}{{{r}^{3}}}\]
D) \[\overrightarrow{dB}=\frac{{{\mu }_{0}}}{4\pi }\frac{\overrightarrow{dl}\times \vec{r}}{{{r}^{4}}}\]
Correct Answer: C
Solution :
If unit current is flowing through the conductor, then Biot-Savarts law is represented as \[\overrightarrow{dB}=\frac{{{\mu }_{0}}}{4\pi }\frac{\overrightarrow{dl}\times \vec{r}}{{{r}^{3}}}\]
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