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Physics
The vector form of Biot-Savart's law for a current carrying element is
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Q. The vector form of Biot-Savart's law for a current carrying element is
Chhattisgarh PMT
Chhattisgarh PMT 2006
A
$d \vec{ B }=\frac{\mu_{0}}{4 \pi} \frac{I d \vec{ l } \sin \phi}{r^{2} }$
B
$d \vec{ B }=\frac{\mu_{0}}{A \pi} \frac{I d l \times \hat{ r }}{r^{2}}$
C
$d \vec{ B }=\frac{\mu_{0}}{4 \pi} \frac{I d \vec{ l } \times \hat{ r }}{r^2}$
D
$d \vec{ B }=\frac{\mu_{0}}{4 \pi} \frac{I d \vec{ l } \times \hat{ r }}{r^2}$
Solution:
Biot- Savart's law, $d B=\frac{\mu_{0}}{4 \pi} \frac{I d l \sin \theta}{r^{2}}$
In vector form, $d \vec{B}=\frac{\mu_{0}}{4 \pi} \frac{I d \vec{l} \times \hat{r}}{r^{2}}$