Question

A current of 5 A is passing along +X direction through a wire lying along X-axis. Find the magnetic field \( \vec{B} \) at a point \( \vec{r} = (3 \hat{i} + 4 \hat{j}) \, \text{m} \) due to 1 cm element of the wire, centered at the origin.

Hint:

The Biot-Savart law is a powerful tool for calculating the magnetic field due to current elements. Remember to carefully compute the cross product between \(d\vec{l}\) and \(\vec{r}\) to find the direction of the magnetic field.

Answer 1

Magnetic Field Due to a Small Current Element Using the Biot-Savart Law 

To find the magnetic field at a point due to a small current element, we use the Biot-Savart law:

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

Where:

• \( d\vec{l} \) is the small element of the wire,
• \( \vec{r} \) is the position vector from the current element to the point where the field is being calculated,
• \( r \) is the distance from the element to the point.

Step 1: Setup and Vector Analysis

Given:

• The current \( I = 5 \, \text{A} \),
• The position vector of the point \( \vec{r} = 3 \hat{i} + 4 \hat{j} \),
• The element of the wire is 1 cm or \( 10^{-2} \, \text{m} \), aligned along the X-axis, so \( d\vec{l} = dx \, \hat{i} \).

The distance \( r \) is:

\[ r = \sqrt{x^2 + 16} \]

Step 2: Applying the Biot-Savart Law

Now, we calculate \( d\vec{B} \):

\[ d\vec{B} = \frac{\mu_0 I}{4 \pi} \frac{dx \, (\hat{i} \times (3 \hat{i} + 4 \hat{j}))}{(x^2 + 16)^{3/2}} \]

The cross product \( \hat{i} \times (3 \hat{i} + 4 \hat{j}) \) simplifies to:

\[ \hat{i} \times (3 \hat{i} + 4 \hat{j}) = 4 \hat{k} \]

Thus,

\[ d\vec{B} = \frac{\mu_0 I}{4 \pi} \frac{4 dx \, \hat{k}}{(x^2 + 16)^{3/2}} \]

Step 3: Integration

To find the total magnetic field, we integrate over the length of the wire element, which extends from \( -0.5 \, \text{cm} \) to \( 0.5 \, \text{cm} \):

\[ B = \int_{-0.5}^{0.5} \frac{\mu_0 I}{4 \pi} \frac{4 dx}{(x^2 + 16)^{3/2}} \]

Step 4: Solving the Integral

On solving, the result gives:

\[ B = 1.6 \times 10^{-10} \, \text{T} \]

Thus, the magnetic field at the point is \( 1.6 \times 10^{-10} \, \text{T} \).