The magnetic field dB→ due to a small current element dl→ at a distance r→ and element carrying current i is

d B → = μ 0 4 π i d l → × r → r 3

d B → = μ 0 4 π i d l → × r → r 3

d B → = μ 0 4 π i 2 d l → × r → r

d B → = μ 0 4 π i d l → × r → r

d B → = μ 0 4 π i 2 d l → × r → r 2

d B → = μ 0 4 π i d l → × r → r 3

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

The magnetic field $d \vec{B}$ due to a small current element $d \vec{l}$ at a distance $\vec{r}$ and element carrying current i is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\vec{d B} = \frac{μ_{0}}{4 π} i^{2} (\frac{d \vec{l} \times \vec{r}}{r})$

$\vec{d B} = \frac{μ_{0}}{4 π} i (\frac{d \vec{l} \times \vec{r}}{r})$

$\vec{d B} = \frac{μ_{0}}{4 π} i^{2} (\frac{d \vec{l} \times \vec{r}}{r^{2}})$

$\vec{d B} = \frac{μ_{0}}{4 π} i (\frac{d \vec{l} \times \vec{r}}{r^{3}})$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

$dB = \frac{μ_{0}}{4 π} . \frac{idl \sin θ}{r^{2}} \Rightarrow d \vec{B} = \frac{μ_{0}}{4 π} . \frac{i (d \vec{l} \times \vec{r})}{r^{3}}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE