Numerical Problems Based on Biot-Savart Law for Class 12 Physics

Given parameters: Current $I = 10\text{ A}$, length of element $dl = 1.8\text{ cm} = 1.8 \times 10^{-2}\text{ m}$, distance $r = 300\text{ cm} = 3\text{ m}$.

The current flows from West to East. The point is North-East, so the angle $\theta$ between the current element and the position vector is $45^\circ$.

According to the Biot-Savart Law:

Substitute the values (where $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T m/A}$):

Applying the Right-Hand Thumb Rule, curling fingers from the East vector towards the North-East vector gives a direction pointing normally out of the plane of paper.

Given current element $I d\vec{l} = 2(2 \times 10^{-3}\hat{k}) = 4 \times 10^{-3}\hat{k}\text{ Am}$.

The vector form of the Biot-Savart Law is $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}$ or $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}$.

(i) At $x = 3\text{ m}$ ($\vec{r} = 3\hat{i}$, $r = 3\text{ m}$):

Since $\hat{k} \times \hat{i} = \hat{j}$:

(ii) At $x = -6\text{ m}$ ($\vec{r} = -6\hat{i}$, $r = 6\text{ m}$):

(iii) At $z = 3\text{ m}$ ($\vec{r} = 3\hat{k}$, $r = 3\text{ m}$):

Here, $d\vec{l}$ is along $\hat{k}$ and $\vec{r}$ is also along $\hat{k}$. The angle between them is $0^\circ$.

Given parameters:
Current $I = 2\text{ A}$.
Element vector $\Delta\vec{l} = \Delta x \hat{i} = 1\text{ cm } \hat{i} = 10^{-2} \hat{i}\text{ m}$.
Position vector of point P on the y-axis, $\vec{r} = 1.0 \hat{j}\text{ m}$. Distance $r = 1.0\text{ m}$.

Using the vector form of Biot-Savart Law:

Substitute the values:

Since the cross product $\hat{i} \times \hat{j} = \hat{k}$:

The magnitude is $2 \times 10^{-9}\text{ T}$ and the direction is along the $+z$-axis.