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:
\[ dB = 4 \times 10^{-8} \hat{k} \, \text{T} \]
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).