Question

Horizontal components of earth at a place is 3.2 ×10^-5 T and angle of dip is 60° . The resultant intensity of earth's field at the place is

Answer 1

Answer:

★ BNET = 6.2 × 10^-5 Tesla ★

Explanation:

★ EARTH MAGNETISM ★

★ GIVEN ;

» Horizontal Magnetic Field ( BH ) = 3.2 × 10^-5 T

» Angle of Dip ( I ) = 60°

★ RESULTANT MAGNETIC FIELD INTENSITY ( Bnet ) = ???

_____ [ BY USING FORMULA ] ______

$\tan(i) \: = \: \frac{BV}{BH}$

★ BV = BH × Tan I

» BV = 3.2 × 10^-5 × Tan 60°

» BV = 3.2 × 10^-5 × √3

» BV = 5.4 × 10^-5

★ VERTICAL COMPONENT OF EARTH'S MAGNETIC FIELD ( BV ) =

5.4 × 10^-5 T ★

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$BNET \: = \: \sqrt{ { {BV}^{2} \: + \: {BH}^{2} \: } }$

★ BNET = √ ( 5.4 × 10^-5 )² + ( 3.2 × 10^-5 )²

» BNET = √ ( 29.16 × 10^-10 + 10.24 × 10^-10

» BNET = √ 39.4 × 10^-10

★ BNET = 6.2 × 10^-5 Tesla ★

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ANSWER :- NET MAGNETIC FIELD OF EARTH HAVING ANGLE OF DIP 60° IS ;

[ BNET = 6.2 × 10^-5 T ]