Question

The vector form of Biot-Savarts law for a current carrying element is

• A. $ d\mathbf{\vec{B}}=\frac{{{\mu }_{0}}}{4\pi }\frac{Id\mathbf{\vec{1}}\,\sin \,\phi }{{{r}^{2}}} $
• B. $ \mathbf{\vec{B}}=\frac{{{\mu }_{0}}}{4\pi }\frac{Idl\,\mathbf{\vec{r}}}{{{r}^{2}}} $
• C. $ d\mathbf{\vec{B}}=\frac{{{\mu }_{0}}}{4\pi }\frac{Id\mathbf{\vec{1}}\times \,\mathbf{\vec{r}}}{{{r}^{3}}} $
• D. $ d\,\mathbf{\vec{B}}=\frac{{{\mu }_{0}}}{4\pi }\frac{Id\mathbf{\vec{1}}\times \,\mathbf{\vec{r}}}{{{r}^{2}}} $

Answer 1

Answer: (D)

Biot-Savaits law, $ dB=\frac{{{\mu }_{0}}}{4\pi }\frac{Idl\sin \theta }{{{r}^{2}}} $ In vector form, $ d\overset{\to }{\mathop{\mathbf{B}}}\,=\frac{{{\mu }_{0}}}{4\pi }\frac{Id\overset{\to }{\mathop{\mathbf{l}}}\,\times \widehat{\mathbf{r}}}{{{r}^{2}}} $