The horizontal component of the earth's magnetic field at a place is 0.25 \\times 10^{-4} {Wb} \/ {m}^{2} and its vertical component is 0.35\\times 10^{-4} {Wb} \/ {m}^{2}. Then the earth's total magnetic field at that place is

Correct Answer is : 0.43×10−4Wb∕m2

0.25 \\times 10^{-4} {Wb} \/ {m}^{2}

0.35 \\times 10^{-4} {Wb} \/ {m}^{2}

0.43\\times 10^{-4} {Wb} \/ {m}^{2}

0.60\\times 10^{-4} {Wb} \/ {m}^{2}

CGPET 2017 Solved Paper

Section: Physics

Question : 46 of 150

Marks: +1, -0

The horizontal component of the earth's magnetic field at a place is 0.25×10−4Wb∕m2 and its vertical component is 0.35×10−4Wb∕m2. Then the earth's total magnetic field at that place is

0.25×10−4Wb∕m2
0.35×10−4Wb∕m2
0.43×10−4Wb∕m2
0.60×10−4Wb∕m2

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