MPSC Rajyaseva 2017 GS Paper 1 — Question 51
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पृथ्वीवरील एका विशिष्ट ठिकाणी चुंबकीय शिखराचा आडवा घटक 0.26 एकक असून तेथील उतार कोन 60° आहे. या ठिकाणी पृथ्वीचे चुंबकीय क्षेत्र किती एकक असेल? In the magnetic meridian of a certain place, the horizontal component of the Earth's magnetic field is 0.26 units and the angle of dip is 60°. What is the magnetic field of the Earth at this location?
1)0.52 एकक / 0.52 units✓ Correct
2)0.13 एकक / 0.13 units
3)15.6 एकक / 15.6 units
4)60.26 एकक / 60.26 units
Using relation B = Bh / cos(theta), where Bh = 0.26 and theta = 60 degrees, total magnetic field B = 0.26 / cos(60) = 0.26 / 0.5 = 0.52 units.