(a) Statement of Biot Savart law Expression in vector form Magnitude of magnetic field at centre Direction of magnetic field(a) It states that magnetic field strength, $\overset{\rightarrow}{d B}$ , due to a current element, $\overset{\rightarrow}{I d l}$ , at a point, having a position vector $r$ relative to the current element, is found to depend (i) directly on the current element, (ii) inversely on the square of the distance $|r|$ , (iii) directly on the sine of angle between the current element and the position vector $r$. In vector notation, $\overset{\rightarrow}{d B}= \; \frac{\mu_0}{4\pi} \; \frac{\overset{\rightarrow}{I d i \times \overset{\rightarrow}{r}}}{\overset{\rightarrow}{|r|^3}}$ (b) $\; B_\pi = \; \frac{\mu_0 \times 1}{2R} = \; \frac{\mu_0}{2R}$ $( \; \text{ along } \; z - \; \text{ direction) } \;$ $\; B_{Q} = \; \frac{\mu_0 \times \sqrt{3}}{2R} = \; \frac{\mu_0 \sqrt{3}}{2R}$ $\; \text{ (along } \; x - \; \text{ direction) } \;$ $\; \therefore B = \sqrt{B_p^2 + B_{Q}^2} = \; \frac{\mu_0}{R}$ This net magnetic field $B$, is inclined to the field $B_{p}$, at an angle $\theta$, where $\; \tan \theta = \sqrt{3}$ $\; ( \theta = \tan^{-1} \sqrt{3} = 60^{\circ} )$ (in $X Z$ plane)