The angle of dip at a place is 30^{circ},and | vedclass

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(B) Given: Angle of dip $\delta = 30^{\circ}$,Horizontal component $B_H = 5 	imes 10^{-5} \, T$.The vertical component $B_V$ is given by the formula:

Vedclass · Accepted Answer

$(5 / \sqrt{3}) 	imes 10^{-5} \, T, (10 / \sqrt{3}) 	imes 10^{-5} \, T$

Vedclass · Answer

(B) Given: Angle of dip $\delta = 30^{\circ}$,Horizontal component $B_H = 5 	imes 10^{-5} \, T$.The vertical component $B_V$ is given by the formula: $B_V = B_H 	an \delta$.Substituting the values: $B_V = 5 	imes 10^{-5} 	imes 	an(30^{\circ}) = 5 	imes 10^{-5} 	imes (1 / \sqrt{3}) = (5 / \sqrt{3}) 	imes 10^{-5} \, T$.The total magnetic field $B$ is given by: $B = B_H / \cos \delta$.Substituting the values: $B = 5 	imes 10^{-5} / \cos(30^{\circ}) = 5 	imes 10^{-5} / (\sqrt{3} / 2) = (10 / \sqrt{3}) 	imes 10^{-5} \, T$.Thus,the vertical component is $(5 / \sqrt{3}) 	imes 10^{-5} \, T$ and the total magnetic field is $(10 / \sqrt{3}) 	imes 10^{-5} \, T$.

The angle of dip at a place is $30^{\circ}$,and the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, T$. Calculate the vertical component and the total magnetic field at that place.

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