The angle of dip at a certain place is 30^o. | vedclass

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(B) The horizontal component of the Earth's magnetic field $(B_H)$ is related to the total magnetic field $(B)$ and the angle of dip $(\phi)$ by the f

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$\frac{2H}{\sqrt{3}}$

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(B) The horizontal component of the Earth's magnetic field $(B_H)$ is related to the total magnetic field $(B)$ and the angle of dip $(\phi)$ by the formula: $B_H = B \cos \phi$.Given that $B_H = H$ and $\phi = 30^o$,we can substitute these values into the equation:$H = B \cos 30^o$Since $\cos 30^o = \frac{\sqrt{3}}{2}$,we have:$H = B 	imes \frac{\sqrt{3}}{2}$Solving for the total magnetic field $B$:$B = \frac{2H}{\sqrt{3}}$Therefore,the intensity of the total magnetic field is $\frac{2H}{\sqrt{3}}$.

The angle of dip at a certain place is $30^o$. If the horizontal component of the earth's magnetic field is $H$,the intensity of the total magnetic field is:

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