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 (1996)

Correct Answer is : d→B=µ04πi(d→l×→rr3)

d \overrightarrow{B}= {\\mu_{0}}/{4 \\pi} i^{2} ( {d {l} ^{\to}\times\overrightarrow{r}}/{r} )

d \overrightarrow{B}= {\\mu_{0}}/{4 \\pi} i ( {d {l} ^{\to}\times\overrightarrow{r}}/{r}^3 )

d \overrightarrow{B}= {\\mu_{0}}/{4 \\pi} i ( {d {l} ^{\to}\times\overrightarrow{r}}/{r} )

d \overrightarrow{B}= {\\mu_{0}}/{4 \\pi} i^{2} ( {d {l} ^{\to}\times\overrightarrow{r}}/{r}2 )

Class 12 NEET Physics Moving Charges and Magnetism

Question : 87 of 100

Marks: +1, -0

The magnetic field d

→

due to a small current element d

→

at a distance

→

and element carrying current i is

(1996)

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