Physics - Magnetism Question with Solution | TestHub

For a dipole at position $(\mathrm{R},\mathrm{Q})$ ${\mathrm{B}}_{\mathrm{R}}=\frac{{\mu }_0}{4\pi }\cdot \frac{2\mathrm{M}\cos \theta }{{\mathrm{R}}^3}$ ...(i) and ${\mathrm{B}}_{\mathrm{Q}}=\frac{{\mu }_0}{4\pi }\cdot \frac{\mathrm{M}\sin \theta }{{\mathrm{R}}^3}$ ...(ii) Also $\tan \varphi =\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}}=-\frac{{\mathrm{B}}_{\mathrm{R}}}{{\mathrm{B}}_{\mathrm{Q}}}$ ...(iii)
Dividing eq. (i) by (ii) $\frac{{\mathrm{B}}_{\mathrm{R}}}{{\mathrm{B}}_{\mathrm{Q}}}=\frac{2\cos \theta }{\sin \theta }=2\cot \theta$ From eq. (iii) and (iv) $\tan \varphi =-2\cot \theta$ From figure, $\theta =90^{\circ }+\lambda$ $\therefore \tan \varphi =-2\cot (90+\lambda )$ $\tan \varphi =2\tan \lambda$