An element `dvecl=dxhati` (where `dx=1cm`) is placed at the origin and carries a large current `i=10A`. What is the magnetic field on the Y-axis at a distance of `0.5m`?

Options

A. `2 xx 10^(-8)hat(k)T`
B. `4 xx 10^(-8)hat(k)T`
C. `-2 xx 10^(-8)hat(k)T`
D. `-4 xx 10^(-8)hat(k)T`

Explanation

To solve the problem, we will use the Biot-Savart Law, which describes the magnetic field generated by a current-carrying conductor. The law states:

\[ \vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \, d\vec{l} \times \vec{r}}{r^3} \]

Where:
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)).
- \( I \) is the current (10 A in this case).
- \( d\vec{l} \) is the differential length element (given as \( dx \hat{i} \)).
- \( \vec{r} \) is the position vector from the current element to the point where we want to find the magnetic field.
- \( r \) is the distance from the current element to the point in question.

### Step-by-Step Solution:

1. **Identify the Parameters:**
- The current \( I = 10 \, \text{A} \).
- The differential length \( d\vec{l} = dx \hat{i} \) where \( dx = 1 \, \text{cm} = 0.01 \, \text{m} \).
- The point where we want to find the magnetic field is at a distance \( r = 0.5 \, \text{m} \) along the y-axis.

2. **Determine the Position Vector \( \vec{r} \):**
- The position vector \( \vec{r} \) from the current element at the origin to the point on the y-axis is \( \vec{r} = 0.5 \hat{j} \).

3. **Calculate the Cross Product \( d\vec{l} \times \vec{r} \):**
- We have \( d\vec{l} = dx \hat{i} \) and \( \vec{r} = 0.5 \hat{j} \).
- The cross product \( d\vec{l} \times \vec{r} = (dx \hat{i}) \times (0.5 \hat{j}) = dx \cdot 0.5 (\hat{i} \times \hat{j}) = 0.5 dx \hat{k} \).

4. **Calculate \( r^3 \):**
- The distance \( r = 0.5 \, \text{m} \), thus \( r^3 = (0.5)^3 = 0.125 \, \text{m}^3 \).

5. **Substitute into the Biot-Savart Law:**
- Now substituting into the formula:
\[ \vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot 0.5 \cdot dx \hat{k}}{r^3} \]
- Substituting the values:
\[ \vec{B} = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{10 \cdot 0.5 \cdot 0.01 \hat{k}}{0.125} \]
- This simplifies to:
\[ \vec{B} = 10^{-7} \cdot \frac{0.05 \hat{k}}{0.125} \]
- Calculating the fraction:
\[ \frac{0.05}{0.125} = 0.4 \]
- Thus,
\[ \vec{B} = 10^{-7} \cdot 0.4 \hat{k} = 4 \times 10^{-8} \hat{k} \, \text{T} \]

6. **Final Result:**
- The magnetic field at the point on the y-axis at a distance of 0.5 m is:
\[ \vec{B} = 4 \times 10^{-8} \hat{k} \, \text{T} \]