Angle of dip at a place is 30^o, and the hori | vedclass

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Question

We know that $\mathrm{B}_{\mathrm{H}}$ $cos$ $\delta=\mathrm{B},$ and $	an \delta=\mathrm{B}_{\mathrm{V}} / \mathrm{B}_{\mathrm{H}}$

Thus $\mathrm

Vedclass · Accepted Answer

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

Vedclass · Answer

We know that $\mathrm{B}_{\mathrm{H}}$ $cos$ $\delta=\mathrm{B},$ and $	an \delta=\mathrm{B}_{\mathrm{V}} / \mathrm{B}_{\mathrm{H}}$

Thus $\mathrm{B}_{\mathrm{v}}=\mathrm{B}_{\mathrm{H}}$ $tan$ $\delta$

${=5 	imes 10^{-5} 	imes 	an \left(30^{\circ}ight)}$

${=(5 / \sqrt{3}) 	imes 10^{-5}}$

and $\mathrm{B}=\mathrm{B}_{\mathrm{H}} / \mathrm{cos} \delta$

${=5 	imes 10^{-5}(\sqrt{3} / 2)}$

${=(10 / \sqrt{3}) 	imes 10^{-5}}$

Angle of dip at a place is $30^o,$ and the horizontal component of earth's field is $5 \times 10^{-5}$ $tesla$, then the vertical component and the total magnetic field at that place is :-

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