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CGPET 2015 Solved Paper

Section: Physics

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Question : 4 of 150

Marks: +1, -0

The horizontal component of earth's magnetic field at a place is

0.4 \times 10^{4} \,\text{T}

. If the angle of dip is

45^{\circ}

the value of total intensity of earth's magnetic field is

$0.5 \times 10^{-4} \,\text{T}$
$0.4 \times 10^{-4} \,\text{T}$
$0.5 \times 10^{-6} \,\text{T}$
$0.4 \times 10^{-6} \,\text{T}$

Solution:

Vertically component of earth's magnetic field is

B_V = B_H \tan \delta

where,

B_H

= horizontal component of earth's magnetic field

\delta = \text{angle of dip}

Total intensity of earth's magnetic field

B = \sqrt{B^6_H + B^2_V}

= \sqrt{B^2_H + B^2_H} [\because \tan \delta = \tan 45^{\circ} = 1]

= \sqrt{2} B_H

= 1.44 \times 0.4 \times 10^{-4} \,\text{T}

= 5.656 \times 10^{-5} \,\text{T}

= 0.5 \times 10^{-4} \,\text{T}

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