Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ph-1741: NEET 2012 Physics - Class 12 AIPM-2012-Ph-1741 Solution: A geostationary satellite is orbiting the earth at a height of…

A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is

5

10

6\sqrt{2}

6/\sqrt{2}

Multiple Choice Options

A ${{\sqrt 5 {\mu _0}I} \over {2R}}$ 2 R 5 _ _ 0 _ I _
B ${{\sqrt 5 {\mu _0}I} \over R}$ R 5 _ _ 0 _ I _
C ${{{\mu _0}I} \over {2R}}$ 2 R _ 0 _ I _
D ${{{\mu _0}I} \over R}$ R _ 0 _ I _

Step-by-Step Verified Solution

Magnetic field induction due to vertical loop at the centre O is B 1 = _ 0 I 2 R It acts in horizontal direction. Magnetic field induction due to horizontal loop at the centre O is B 2 = _ 0 2 I 2 R It acts in vertically upward direction. As B 1 and B 2 are perpendicular to each other, therefore the resultant magnetic field induction at the centre O is B n e t = B 1 2 + B 2 2 = ( _ 0 I 2 R ) 2 + ( _ 0 2 I 2 R ) 2 B n e t = _ 0 I 2 R ( 1 ) 2 + ( 2 ) 2 = 5 _ 0 I 2 R

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