The horizontal component of the electric field is given by BH = Becos(θ)where θ is the angle of dip at the given placeBe is teh net magnetic fieldAt θ = 60oBH = Be cos(60) = Be x (1/2)Be = 2BH = 2B (given BH =B)Now horizontal component at equatorBH = Becos (θ)angle of dip at equator is 0BH = Becos(0) = Be = 2BBH = 2B
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