Question

At a place horizontal component of the earth's magnetic field is B and angle of dip at the place is 60^o. What is the value of horizontal component of the earth's magnetic field at a place where the dip angle is 30^o ?

Answer 1

$$\begin{gathered} \mathrm{Horzontal}\mathrm{component}\mathrm{of}\mathrm{magnetic}\mathrm{field}, \\ {\mathrm{B}}_{\mathrm{H}}=\mathrm{B}\text{'}\mathrm{cos\delta } \\ \mathrm{According}\mathrm{to}\mathrm{question}, \\ \mathrm{B}=\mathrm{B}\text{'}\cos {60}^{}=\mathrm{B}\text{'}/2 \\ \mathrm{So},\mathrm{earth}\text{'}\mathrm{s}\mathrm{magnetic}\mathrm{field}\mathrm{intensity},\mathrm{B}\text{'}=2\mathrm{B} \\ \mathrm{Now}\mathrm{again}, \\ \mathrm{\delta }={30}^0 \\ \mathrm{So}, \\ {\mathrm{B}}_{\mathrm{H}}=2\mathrm{Bcos}{30}^0=2\mathrm{B}\times \frac{\sqrt{3}}{2}=\sqrt{3}\mathrm{B} \end{gathered}$$