The vector form of Biost-Savart law for a current carrying element is

Options

A. `d vec(B) = (mu_(0))/(4pi) (idl sin phi)/(r^(2))`
B. `d vec(B) = (mu_(0))/(4pi) (ivec(dl) xx hat(r))/(r^(2))`
C. `d vec(B) = (mu_(0))/(4pi) (ivec(dl) xx hat(r))/(r^(3))`
D. `dvec(B) = (mu_(0))/(4pi) (ivec(dl)xx vec(r))/(r^(2))`

Answer

To derive the vector form of the Biot-Savart law for a current-carrying element, we can follow these steps:

### Step 1: Understand the Concept
The Biot-Savart law describes the magnetic field generated by a current-carrying conductor. It states that the magnetic field \( \mathbf{B} \) at a point in space due to a small segment of current-carrying wire is proportional to the current, the length of the wire segment, and the sine of the angle between the wire segment and the line connecting the wire segment to the point.

### Step 2: Define the Variables
Let:
- \( I \) = current flowing through the wire
- \( d\mathbf{l} \) = infinitesimal length vector of the wire segment
- \( \mathbf{r} \) = position vector from the wire segment to the point where the magnetic field is being calculated
- \( r \) = magnitude of vector \( \mathbf{r} \)
- \( \hat{\mathbf{r}} \) = unit vector in the direction of \( \mathbf{r} \) (i.e., \( \hat{\mathbf{r}} = \frac{\mathbf{r}}{r} \))

### Step 3: Write the Biot-Savart Law
The magnetic field \( d\mathbf{B} \) due to the current element \( d\mathbf{l} \) is given by:
\[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \hat{\mathbf{r}}}{r^2} \]
where:
- \( \mu_0 \) = permeability of free space
- \( \times \) = cross product

### Step 4: Substitute the Unit Vector
The unit vector \( \hat{\mathbf{r}} \) can be expressed as:
\[ \hat{\mathbf{r}} = \frac{\mathbf{r}}{r} \]
Substituting this into the equation gives:
\[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \frac{\mathbf{r}}{r}}{r^2} \]
This simplifies to:
\[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \mathbf{r}}{r^3} \]

### Step 5: Final Form
Thus, the vector form of the Biot-Savart law for a current-carrying element is:
\[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \mathbf{r}}{r^3} \]

### Conclusion
The correct expression for the vector form of the Biot-Savart law is:
\[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \hat{\mathbf{r}}}{r^2} \]