Question

If \(F\) represents the Earth's total magnetic field corresponding to its dipole source only, then the increase in \(F\) from 0\(^\circ\) to 60\(^\circ\) N magnetic latitude is __________ % (rounded off to one decimal place).

Hint:

Use the geocentric dipole field formula \(F(\lambda) = F_0\sqrt{1+3\sin^2\lambda}\) and compare its values at 0 and 60 degrees.

Answer 1

Correct Answer: 80

Instead of quoting the combined dipole formula, build \(F\) up from its two field components and reach the same number a different way.

Step 1: Horizontal and vertical dipole components.
\(H(\lambda) = F_0\cos\lambda\), \(Z(\lambda) = 2F_0\sin\lambda\).

Step 2: Total field.
\(F(\lambda) = \sqrt{H^2+Z^2} = F_0\sqrt{\cos^2\lambda + 4\sin^2\lambda} = F_0\sqrt{1+3\sin^2\lambda}\) (using \(\cos^2\lambda = 1-\sin^2\lambda\)), the identical dipole relation.

Step 3: Evaluate at 60\(^\circ\).
\(H = F_0\cos60^\circ = 0.5F_0\), \(Z = 2F_0\sin60^\circ = 1.7321F_0\).
\(F(60^\circ) = F_0\sqrt{0.25+3.0} = F_0\sqrt{3.25} = 1.8028F_0\).

Percentage increase over the equatorial value \(F_0\) is \((1.8028-1)\times100 \approx \boxed{80.3\%}\), matching the direct-formula result.