The magnetic field `bar(dB)` due to a small current element `bar(dl)` at a distance `vec(r)` and carrying current `'I'` is

Options

A. `dB=(mu_(0))/(4pi)i((dlxxr)/(r))`
B. `dB=(mu_(0))/(4pi)i^(2)((dlxxr)/(r^(2)))`
C. `dB=(mu_(0))/(4pi)i^(2)((dlxxr)/(r))`
D. `dB=(mu_(0))/(4pi)i((dlxxr)/(r^(3)))`

Answer

To find the magnetic field \( \bar{dB} \) due to a small current element \( \bar{dl} \) at a distance \( \vec{r} \) carrying current \( I \), we will use the Biot-Savart Law. The steps to derive the expression for the magnetic field are as follows:

### Step-by-Step Solution:

1. **Understanding the Biot-Savart Law**:
The Biot-Savart Law states that the magnetic field \( \bar{dB} \) at a point in space due to a small current element \( \bar{dl} \) carrying current \( I \) is directly proportional to the current, the length of the current element, and the sine of the angle between the current element and the line connecting the current element to the point where the magnetic field is being calculated.

2. **Formula for the Magnetic Field**:
The mathematical expression for the Biot-Savart Law is given by:
\[ \bar{dB} = \frac{\mu_0}{4\pi} \frac{I \, \bar{dl} \times \vec{r}}{r^3} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current through the element \( \bar{dl} \),
- \( \vec{r} \) is the position vector from the current element to the point where the magnetic field is being calculated,
- \( r \) is the magnitude of the vector \( \vec{r} \).

3. **Cross Product**:
The term \( \bar{dl} \times \vec{r} \) indicates that the magnetic field direction is perpendicular to both the current element and the position vector. The cross product ensures that the direction of the magnetic field follows the right-hand rule.

4. **Magnitude of \( \vec{r} \)**:
The magnitude \( r \) is calculated as:
\[ r = |\vec{r}| \]

5. **Final Expression**:
Therefore, the expression for the magnetic field \( \bar{dB} \) due to the small current element \( \bar{dl} \) at a distance \( \vec{r} \) is:
\[ \bar{dB} = \frac{\mu_0}{4\pi} \frac{I \, \bar{dl} \times \vec{r}}{r^3} \]