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) \[d\,\mathbf{\vec{B}}=\frac{{{\mu }_{0}}}{4\pi }\frac{Id\mathbf{\vec{1}}\times \,\mathbf{\vec{r}}}{{{r}^{2}}}\]
Correct Answer: D
Solution :
Biot-Savaits law, \[dB=\frac{{{\mu }_{0}}}{4\pi }\frac{Idl\sin \theta }{{{r}^{2}}}\] In vector form, \[d\overset{\to }{\mathop{\mathbf{B}}}\,=\frac{{{\mu }_{0}}}{4\pi }\frac{Id\overset{\to }{\mathop{\mathbf{l}}}\,\times \widehat{\mathbf{r}}}{{{r}^{2}}}\]
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