Question

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its North tip down at 30° with the horizontal.

The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G.

Determine the magnitude of the earth’s magnetic field at the place.


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Answer 1

Explanation:

$Angle of dip, θ=60o</p><p>H=0.4 G=0.4×10−4T</p><p>If Be is earth's magnetic field, then</p><p>H=Becosθ.⇒Be=cosθH=cos60o0.4×10−4T=0.50.4×10−4T=0.8×10−4T=0.8 G</p><p></p><p>$

Answer 2

Answer:

Horizontal component of earth's magnetic field, BH =0.4 G

Angle made by the needle with the horizontal plane = Angle of dip = δ= 600

Earth's magnetic field strength = B

We can relate B and BH as:

B H = B cos ø

B = B H/ cos ø

= 0.4/0.5 = 0.8 G

Hence, the strength of earth's magnetic field at the given location is 0.8G.