The Biot-Savart's law in vector from is:
- $ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { di\left( \overrightarrow { l } \times \overrightarrow { r } \right) }{ r^ 2 } $
- $ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ r^ 2 } $
- $ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { r } \times \overrightarrow { dl } \right) }{ r^ 2 } $
- $ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ r^ 3 } $
Reveal answer
Fill a bubble to check yourself
D
Correct answer
Explanation
Biot-Savart law in vector form gives the magnetic field dB due to a current element idL at position r. The correct expression has the cross product of current element vector dl and position vector r in the numerator, divided by r cubed in the denominator to make it dimensionally consistent as an inverse-square law in vector form.