Horizontal component of the earth-s field is 3 - 10-5 T and of dip is 53o- Magnetic field of the earth at that place is

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Standard XII

Physics

Earth's Magnetism

Horizontal co...

Question

Horizontal component of the earth's field is $3 \times 10^{- 5} T$ and of dip is $53^{o}$ . Magnetic field of the earth at that place is

25 \times 10^{- 5} T

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5 \times 10^{- 5} T

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85 \times 10^{10} T

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9.9 \times 10^{- 8} T

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Solution

The correct option is B $5 \times 10^{- 5} T$
Horizontal component of the earths magnetic fields is given by
$B_{H} = B c o s δ$
$\Rightarrow B = \frac{B_{H}}{c o s δ} = \frac{3 \times 10^{- 5}}{c o s 53^{o}}$
$= 5 \times 10^{- 5} T$

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Horizontal component of the earth's field is 3×10−5T and of dip is 53o. Magnetic field of the earth at that place is

The correct option is B 5×10−5THorizontal component of the earths magnetic fields is given byBH=Bcosδ⇒B=BHcosδ=3×10−5cos53o=5×10−5T

Horizontal component of the earth's field is $3 \times 10^{- 5} T$ and of dip is $53^{o}$ . Magnetic field of the earth at that place is

The correct option is B $5 \times 10^{- 5} T$
Horizontal component of the earths magnetic fields is given by
$B_{H} = B c o s δ$
$\Rightarrow B = \frac{B_{H}}{c o s δ} = \frac{3 \times 10^{- 5}}{c o s 53^{o}}$
$= 5 \times 10^{- 5} T$