The value of the horizontal component of the | vedclass

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(A) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \phi$,where $B$ is the total intensity and $\phi$ is the angle of

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$2.08 	imes 10^{-5} \, Wb/m^2$

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(A) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \phi$,where $B$ is the total intensity and $\phi$ is the angle of dip.Given: $B_H = 1.8 	imes 10^{-5} \, Wb/m^2$ and $\phi = 30^{\circ}$.We need to find the total intensity $B = \frac{B_H}{\cos \phi}$.Substituting the values: $B = \frac{1.8 	imes 10^{-5}}{\cos 30^{\circ}}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$,we have $B = \frac{1.8 	imes 10^{-5}}{0.866} \approx 2.08 	imes 10^{-5} \, Wb/m^2$.

The value of the horizontal component of the earth's magnetic field and the angle of dip are $1.8 \times 10^{-5} \, Wb/m^2$ and $30^{\circ}$ respectively at a certain place. The total intensity of the earth's magnetic field at that place will be:

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