Solution:

Ans. (i) Statement of Biot-Savart’s law

(ii) Magnitude of magnetic field at centre Direction of magnetic field

(i) It states that magnetic field strength, $\overrightarrow{d B}$ due to a current element, $I \vec{d}$ at a point, having a position vector $\vec{r}$ relative to the current element, is found to depend (i) directly on the current element, (ii) inversely on the square of the distance $|\vec{r}|$ (iii) directly on the sine of angle between the current element and the position vector $\vec{r}$.

In vector notation,

$$ \overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \frac{I \overrightarrow{d l} \times \vec{r}}{|\vec{r}|^{3}} $$

Alternatively,

$$ \left(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \cdot \frac{I \overrightarrow{d l} \times \hat{r}}{|\vec{r}|^{2}}\right) $$

(ii)

$$ \begin{aligned} B_{P} & =\frac{\mu_{0} \times 1}{2 R}=\frac{\mu_{0}}{2 R} \ B_{Q} & =\frac{\mu_{0} \times \sqrt{3}}{2 R}=\frac{\mu_{0} \sqrt{3}}{2 R} \ B & =\sqrt{B_{P}^{2}+B_{Q}^{2}}=\frac{\mu_{0}}{R} \end{aligned} $$

(along Z-direction) $1 / 2$

(along $X$-direction) $1 / 2$

$\therefore$

This net magnetic field $B$, is inclined to the field $B_{P}$, at an angle $\theta$ where

$$ \begin{aligned} & \tan \theta=\sqrt{3} \ & \left(\text { or } \theta=\tan ^{-1} \sqrt{3}=60^{\circ}\right) \end{aligned} $$

(in XZ plane) $1 / 2$

OR

Biot-Savart Law

Beot Bavast’s haw stakes thal the

magnetic fceld due to a cuseent eteroonc dl

a) adestance $\vec{x}$. from it is gever by

$\sqrt{d \beta} \alpha d B \propto I$ dBad $d B \alpha \operatorname{sen} \theta$

10 vectoe from

$\overrightarrow{d B}=\frac{140}{4 \pi} I \frac{\left(\pi^{\top} \times \vec{r}\right)}{x^{3}}$

Beal saciast’s law geves the magnetic fold due to a culenc element $a t$ a porncrdestance $\because$

The magnetic fieid die to ebe coil p at the centre

[Topper’s Answer 2017]