A magnetic needle free to rotate in a vertica | vedclass

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Question

(A) The angle of dip $\delta$ is the angle that the total magnetic field of the earth makes with the horizontal direction. Here,$\delta = 30^{\circ}$.

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{5} \ G$

Vedclass · Answer

(A) The angle of dip $\delta$ is the angle that the total magnetic field of the earth makes with the horizontal direction. Here,$\delta = 30^{\circ}$.Given,the horizontal component of the earth's magnetic field is $B_H = 0.3 \ G$.The relationship between the total magnetic field $B$,the horizontal component $B_H$,and the angle of dip $\delta$ is given by $B_H = B \cos \delta$.Substituting the given values: $0.3 = B \cos 30^{\circ}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $0.3 = B 	imes \frac{\sqrt{3}}{2}$.Solving for $B$: $B = \frac{0.3 	imes 2}{\sqrt{3}} = \frac{0.6}{\sqrt{3}} = \frac{0.6 	imes \sqrt{3}}{3} = 0.2 \sqrt{3} \ G$.Alternatively,$B = \frac{0.6}{\sqrt{3}} = \frac{6}{10 \sqrt{3}} = \frac{3}{5 \sqrt{3}} = \frac{\sqrt{3}}{5} \ G$.

Similar Questions

If the magnetic declination at a place is known to be $15^{\circ} \, E$,and a compass needle points as shown in the figure,then the geographic north is represented by which numbered direction?

$A$ magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{\circ}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$ with the horizontal,the true dip value at that place is: (in $^{\circ}$)

What is the angle between the axis of rotation and the magnetic axis of the Earth (in $°$)?

At a point on the surface of the Earth,the value of the horizontal component of the Earth's magnetic field is equal to the value of the vertical component of the Earth's magnetic field. The angle of dip is:

Assertion : The true geographic north direction is found by using a compass needle.
Reason : The magnetic meridian of the earth is along the axis of rotation of the earth.

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