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

Field point position vector:

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

Biot-Savart law for magnetic field from current element $\Delta l$:

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

Where:

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

Cross product calculation:

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

Substitution into Biot-Savart law:

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

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

Resulting magnetic field:

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