VITEEE 2015 Paper

Due to current through side AB Magnetic field at the centre O $B_1$ = $\frac{\mu_0I}{4\pi a}$ $[\sin\theta_1+\sin\theta_2]$ As the magnetic field due to each of the 24. For a dipole at position (R, Q) three sides is the same in magnitude and direction. ∴ Total magnetic field at O is sum of all the fields. i.e. B = $3B_1$ = $\frac{3\mu_0I}{4\pi a}\,[\sin\theta_1+\sin\theta_2]$ Here, tan $\theta_1$ = $\frac{AD}{OD}$ ⇒ tan 60° = $\frac{\frac{l}{2}}{a}$ ⇒ a = $\frac{l}{2\sqrt{3}}$ = $\frac{9\times10^{-2}}{2\sqrt{3}}$ Now B = 3 × $\frac{4\pi\times10^{-7}\times2}{4\pi\times\frac{9\times10^{-2}}{2\sqrt{3}}}$ [sin 60° + sin 60°] = $\frac{4\sqrt{3}}{9}$ × $10^{-5}$ $\left[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]$ 1.33 x $10^{-5}$ T