The vector form of Biot-Savart's law for a current carrying element is

Question

The vector form of Biot-Savart's law for a current carrying element is (A) d vec B =(μ0/4 π) (I d vec l sin φ/r2 ) (B) d vec B =(μ0/A π) (I d l × hat r /r2) (C) d vec B =(μ0/4 π) (I d vec l × hat r /r2) (D) d vec B =(μ0/4 π) (I d vec l × hat r /r2)

Tardigrade Admin · Accepted Answer

Biot- Savart's law, d B=(μ0/4 π) (I d l sin θ/r2) In vector form, d vecB=(μ0/4 π) (I d vecl × hatr/r2)

Q. The vector form of Biot-Savart's law for a current carrying element is

$d B = \frac{μ _{0}}{4 π} \frac{I d l s i n ϕ}{r ^{2}}$

$d B = \frac{μ _{0}}{A π} \frac{I d l \times r ^}{r ^{2}}$

$d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$

$d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$

Biot- Savart's law, $d B = \frac{μ _{0}}{4 π} \frac{I d l s i n θ}{r ^{2}}$
In vector form, $d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$

Q. The vector form of Biot-Savart's law for a current carrying element is

dB=4πμ0​​r2Idlsinϕ​

dB=Aπμ0​​r2Idl×r^​

dB=4πμ0​​r2Idl×r^​

dB=4πμ0​​r2Idl×r^​

Biot- Savart's law, dB=4πμ0​​r2Idlsinθ​ In vector form, dB=4πμ0​​r2Idl×r^​

$d B = \frac{μ _{0}}{4 π} \frac{I d l s i n ϕ}{r ^{2}}$

$d B = \frac{μ _{0}}{A π} \frac{I d l \times r ^}{r ^{2}}$

$d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$

$d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$

Biot- Savart's law, $d B = \frac{μ _{0}}{4 π} \frac{I d l s i n θ}{r ^{2}}$
In vector form, $d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$