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(B) Let $B$ be the total intensity of the Earth's magnetic field at the given place.Let $B_H$ be the horizontal component of the Earth's magnetic fiel

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$\sqrt{2} B_0$

Vedclass · Answer

(B) Let $B$ be the total intensity of the Earth's magnetic field at the given place.Let $B_H$ be the horizontal component of the Earth's magnetic field.Let $	heta$ be the angle of dip.We know that the horizontal component is given by $B_H = B \cos 	heta$.Given $B_H = B_0$ and $	heta = 45^o$.Substituting these values into the formula: $B_0 = B \cos 45^o$.Since $\cos 45^o = \frac{1}{\sqrt{2}}$,we have $B_0 = B 	imes \frac{1}{\sqrt{2}}$.Therefore,$B = \sqrt{2} B_0$.

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 field at that place will be:

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