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  • At a certain place of on earth a magnetic needle is placed along the magnetic meridian at an angle of 60 degree to the horizontal. If the horizontal component of the earth's field at the place

At a certain place of on earth a magnetic needle is placed along the magnetic meridian at an angle of 60 degree to the horizontal. If the horizontal component of the earth's field at the place is  0.20 \times 10^{-4} \mathrm{~T} , what is the magnitude of the total earth's field at that place

Option: 1

0.2 \times 10^{-4} \mathrm{~T}


Option: 2

0.4 \times 10^{-4} \mathrm{~T}


Option: 3

0.8 \times 10^{-4} \mathrm{~T}


Option: 4

1.6 \times 10^{-4} \mathrm{~T}


Answers (1)

best_answer

\mathrm{B_H=0.20 \times 10^{-4} \mathrm{~T}}  and  \mathrm{\theta=60^{\circ}} . We know that  \mathrm{B_H= B \cos \theta}  where B is the magnitude of the total earth's field. Thus,
\mathrm{ B=\frac{B_H}{\cos \theta}=\frac{0.20 \times 10^{-4}}{\cos 60^{\circ}}=0.40 \times 10^{-4} \mathrm{~T} }

Posted by

Pankaj Sanodiya

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