A small current element of length 'd l' and carrying current is placed at (1,1,0) and is carrying current in '+z' direction. If magnetic field at origin be `vecB_(1)` and at point (2,2,0) be `vecB_(2)` then :

Options

A. `|vecB_(1)|=|vecB_(2)|`
B. `vecB_(1)=+vecB_(2)`
C. `|vecB_(1)|=|2vecB_(2)|`
D. `vecB_(1)=-2vecB_(2)`

Explanation

To solve the problem, we need to determine the magnetic fields at two different points due to a small current element. Let's break down the solution step by step.

### Step 1: Understand the Position of the Current Element
The current element of length \( d\ell \) is located at the point \( (1, 1, 0) \) and is carrying current in the +z direction.

### Step 2: Identify the Points of Interest
We need to find the magnetic field at two points:
- At the origin \( (0, 0, 0) \), denoted as \( \vec{B}_1 \)
- At the point \( (2, 2, 0) \), denoted as \( \vec{B}_2 \)

### Step 3: Use the Biot-Savart Law
The magnetic field \( \vec{B} \) due to a current element \( I d\ell \) at a point in space is given by the Biot-Savart Law:

\[ \vec{B} = \frac{\mu_0}{4\pi} \frac{I d\ell \times \hat{r}}{r^2} \]

where:
- \( \mu_0 \) is the permeability of free space,
- \( \hat{r} \) is the unit vector from the current element to the point where the magnetic field is being calculated,
- \( r \) is the distance from the current element to the point.

### Step 4: Calculate the Magnetic Field at the Origin \( \vec{B}_1 \)
1. **Position Vector**: The position vector from the current element at \( (1, 1, 0) \) to the origin \( (0, 0, 0) \) is:
\[ \vec{r}_1 = (0 - 1, 0 - 1, 0 - 0) = (-1, -1, 0) \]
2. **Magnitude of \( r_1 \)**:
\[ r_1 = \sqrt{(-1)^2 + (-1)^2 + 0^2} = \sqrt{2} \]
3. **Unit Vector \( \hat{r}_1 \)**:
\[ \hat{r}_1 = \frac{\vec{r}_1}{r_1} = \frac{(-1, -1, 0)}{\sqrt{2}} = \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right) \]
4. **Cross Product**:
The current element \( d\ell \) is in the +z direction, so \( d\ell = (0, 0, d\ell) \).
\[ I d\ell \times \hat{r}_1 = (0, 0, I d\ell) \times \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right) = \left(I d\ell \frac{1}{\sqrt{2}}, -I d\ell \frac{1}{\sqrt{2}}, 0\right) \]
5. **Magnetic Field \( \vec{B}_1 \)**:
\[ \vec{B}_1 = \frac{\mu_0 I d\ell}{4\pi} \frac{\left(I d\ell \frac{1}{\sqrt{2}}, -I d\ell \frac{1}{\sqrt{2}}, 0\right)}{2} = \frac{\mu_0 I d\ell}{8\pi} \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right) \]

### Step 5: Calculate the Magnetic Field at Point \( (2, 2, 0) \) \( \vec{B}_2 \)
1. **Position Vector**: The position vector from the current element at \( (1, 1, 0) \) to the point \( (2, 2, 0) \) is:
\[ \vec{r}_2 = (2 - 1, 2 - 1, 0 - 0) = (1, 1, 0) \]
2. **Magnitude of \( r_2 \)**:
\[ r_2 = \sqrt{(1)^2 + (1)^2 + 0^2} = \sqrt{2} \]
3. **Unit Vector \( \hat{r}_2 \)**:
\[ \hat{r}_2 = \frac{\vec{r}_2}{r_2} = \frac{(1, 1, 0)}{\sqrt{2}} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \]
4. **Cross Product**:
\[ I d\ell \times \hat{r}_2 = (0, 0, I d\ell) \times \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) = \left(-I d\ell \frac{1}{\sqrt{2}}, I d\ell \frac{1}{\sqrt{2}}, 0\right) \]
5. **Magnetic Field \( \vec{B}_2 \)**:
\[ \vec{B}_2 = \frac{\mu_0 I d\ell}{4\pi} \frac{\left(-I d\ell \frac{1}{\sqrt{2}}, I d\ell \frac{1}{\sqrt{2}}, 0\right)}{2} = \frac{\mu_0 I d\ell}{8\pi} \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \]

### Step 6: Compare \( \vec{B}_1 \) and \( \vec{B}_2 \)
From the calculations:
- \( \vec{B}_1 = \frac{\mu_0 I d\ell}{8\pi} \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right) \)
- \( \vec{B}_2 = \frac{\mu_0 I d\ell}{8\pi} \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \)

### Conclusion
The magnitudes of \( \vec{B}_1 \) and \( \vec{B}_2 \) are the same, but their directions are opposite. Thus, we can conclude that:

\[ \vec{B}_1 = -\vec{B}_2 \]