JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Question Bank Biot-savart's Law and Amperes Law

  • question_answer
    The magnetic field \[d\overrightarrow{B}\] due to a small current element \[d\overrightarrow{l\,}\] at a distance \[\overrightarrow{r\,}\]  and element carrying current i is, or            Vector form of Biot-savart's law is  [CBSE PMT 1996; MP PET 2002; MP PMT 2000]

    A)            \[d\overrightarrow{B}=\frac{{{\mu }_{0}}}{4\pi }i\,\left( \frac{d\overrightarrow{l\,}\times \overrightarrow{r\,}}{r} \right)\]    

    B)            \[d\overrightarrow{B}=\frac{{{\mu }_{0}}}{4\pi }{{i}^{2}}\,\left( \frac{d\overrightarrow{l\,}\times \overrightarrow{r\,}}{r} \right)\]

    C)            \[d\overrightarrow{B}=\frac{{{\mu }_{0}}}{4\pi }{{i}^{2}}\,\left( \frac{d\overrightarrow{l\,}\times \overrightarrow{r\,}}{{{r}^{2}}} \right)\]                                 

    D)            \[d\overrightarrow{B}=\frac{{{\mu }_{0}}}{4\pi }i\,\left( \frac{d\overrightarrow{l\,}\times \overrightarrow{r\,}}{{{r}^{3}}} \right)\]

    Correct Answer: D

    Solution :

               \[dB=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{idl\sin \theta }{{{r}^{2}}}\] Þ \[d\overrightarrow{B}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{i\,(d\overrightarrow{l\,}\times \overrightarrow{r\,})}{{{r}^{3}}}\]


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