A long curved conductor carries a current I (I is a vector). A small current element of length $\mathbf{d}…

A long curved conductor carries a current I (I is a vector). A small current element of length $\mathbf{d} l$, on the wire induces a magnetic field at a point, away from the current element. If the position vector between the current element and the point is $\mathbf{r}$, making an angle with current element then, the induced magnetic field density; $\mathbf{d B}$ (vector) at the point is ( $\mu_0=$ permeability of free space)
  1. $\frac{\mu_0 \mathbf{I d} l \times \mathbf{r}}{4 \pi r}$ (perpendicular to the current element $\mathbf{d} l$ )
  2. $\frac{\mu_0 \mathbf{I} \times \mathbf{r} \times \mathbf{d} l}{4 \pi r^2}$ (perpendicular to the current element $\mathbf{d} l$ )
  3. $\frac{\mu_0 \mathbf{I} \times \mathbf{d} l}{r}$ (perpendicular to the plane containing the current element and position vector $\mathbf{r}$ )
  4. $\frac{\mu_0 \mathbf{I} \times \mathbf{d} l}{4 \pi r^2}$ (perpendicular to the plane containing current element and position vector $\mathbf{r}$ )

Solution

The magnetic field $d B=\frac{\mu_0 \mathbf{I} \times \mathbf{r} \times \mathbf{d} l}{4 \pi r^2}$

Asked in: AP EAMCET 2012

Practice more Magnetic Effects of Current questions on Aicharya