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(B) The horizontal component of the Earth's magnetic field is given by $B_H = B \cos \delta$,where $B$ is the total intensity and $\delta$ is the angl

Vedclass · Accepted Answer

$\sqrt{2} B_0$

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(B) The horizontal component of the Earth's magnetic field is given by $B_H = B \cos \delta$,where $B$ is the total intensity and $\delta$ is the angle of dip.Given $B_H = B_0$ and $\delta = 45^o$.Substituting the values,we get $B_0 = B \cos 45^o$.Since $\cos 45^o = \frac{1}{\sqrt{2}}$,we have $B_0 = B 	imes \frac{1}{\sqrt{2}}$.Therefore,the total intensity $B = \sqrt{2} B_0$.

At a certain place,the horizontal component of the Earth's magnetic field is $B_0$ and the angle of dip is $45^o$. The total intensity of the Earth's magnetic field at that place will be:

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