$\; B_H = B_E \cos \delta$ $\; B_H = B_E \cos 60^{\circ} \Rightarrow B_E = 2 B_H$ At equator $\delta = 0^{\circ}$ $\therefore \; \; B_H = 2 B \cos 0^{\circ} = 2 B$ [Alternatively, award full one mark, if student doesn't take the value $(=2 B)$ of $B_E$ while finding the value of horizontal component at equator, and just writes the formula only.]