Question

An element \(\triangle l=\triangle x \hat i\) is placed at the origin and carries a large current I = 10A. The magnetic field on the y-axis at a distance of 0.5 m from the elements x of 1 cm length is :

• A. $4 \times 10^{-8} , \text{T}$
• B. $8 \times 10^{-8} , \text{T}$
• C. $12 \times 10^{-8} , \text{T}$
• D. $10 \times 10^{-8} , \text{T}$

Answer 1

The Correct Option is A

The position vector for the field point is:

\[ \vec{r} = 0.5\hat{j} \]

The magnetic field produced by the current element $\Delta l$ is given by the Biot-Savart law:

\[ dB = \frac{\mu_0 I (\Delta \vec{l} \times \vec{r})}{4\pi r^3} \]

Here,

\[ \Delta \vec{l} = \Delta x \hat{i} = \frac{1}{100}\hat{i} \, \text{m}, \quad \vec{r} = 0.5\hat{j} \, \text{m}, \quad r = |\vec{r}| = 0.5 \, \text{m} \]

\[ \Delta \vec{l} \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{100} & 0 & 0 \\ 0 & 0.5 & 0 \end{vmatrix} = \frac{1}{100} \times 0.5 \hat{k} = \frac{1}{200} \hat{k} \]

Substituting values into the Biot-Savart law:

\[ dB = \frac{10^{-7} \times 10 \times \frac{1}{200}}{(0.5)^3} \, \text{T} \]

\[ dB = \frac{10^{-7} \times 10 \times \frac{1}{200}}{\frac{1}{8}} = 4 \times 10^{-8} \, \text{T} \]

Hence, the magnetic field is:

\[ dB = 4 \times 10^{-8} \hat{k} \, \text{T} \]

Answer 2

To solve this problem, we will use the Biot-Savart Law, which gives us the magnetic field \( \text{d}\mathbf{B} \) at a point due to a small current element \( \text{d}\mathbf{l} \). The formula is:

\(\text{d}\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, \text{d}\mathbf{l} \times \mathbf{r}}{r^3}\)

Where:

• \( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \) is the permeability of free space.
• \( I = 10 \text{ A} \) is the current through the element.
• \( \text{d}\mathbf{l} = \triangle x \hat{i} \) is the small element of the wire.
• \( \mathbf{r} \) is the position vector from the element to the point where the magnetic field is being calculated.

Here, \( \triangle x = 0.01 \text{ m} \) and the point is on the y-axis at \( \mathbf{r} = 0.5 \text{ m} \, \hat{j} \). The cross product \( \text{d}\mathbf{l} \times \mathbf{r} = \triangle x \hat{i} \times 0.5 \hat{j} = \triangle x \cdot 0.5 \cdot \hat{k} \).

The magnitude of the position vector is \( r = 0.5 \text{ m} \). Thus, \( r^3 = (0.5)^3 = 0.125 \text{ m}^3 \).

Now, substituting into the Biot-Savart Law, we get:

\(\text{d}B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{10 \times 0.01 \times 0.5}{0.125}\)

Simplifying,

\(\text{d}B = 10^{-7} \cdot \frac{0.05}{0.125} = 4 \times 10^{-8} \text{ T}\)

Therefore, the magnetic field at the given point is:

Option A: \(4 \times 10^{-8} \text{ T}\)