Horizontal component of the earth-s field is 3 times 10-5- - T and of dip is 53-o- Magnetic field of the earth that place is25 times 10-5- - T5 times 10-5- - T85 times 10-10- - T9-9 times 10-8- - T

Question

Toppr Admin · Accepted Answer

The correct option is B 5-xD7-10-x2212-5THorizontal component of the earths magnetic fields is given byBH-Bcos-x3B4-x21D2-B-BHcos-x3B4-3-xD7-10-x2212-5cos53o-5-xD7-10-x2212-5T

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
$25 \times 10^{- 5} T$
$5 \times 10^{- 5} T$
$85 \times 10^{10} T$
$9.9 \times 10^{- 8} T$

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

Horizontal component of the earth's field is 3×10−5T and of dip is 53o. Magnetic field of the earth at that place is25×10−5T5×10−5T85×1010T9.9×10−8T

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
$25 \times 10^{- 5} T$
$5 \times 10^{- 5} T$
$85 \times 10^{10} T$
$9.9 \times 10^{- 8} T$

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