The angle of dip at a place is$37^\circ $ and the vertical component of the earth’s magnetic field is $6 \times {10^{ - 5}}{\rm{\;T}}.$ The earth’s magnetic field at this place is $({\rm{tan\;}}37^\circ = 3/4$)

$7 \times {10^{ - 5}}{\rm{\;T}}$

$6 \times {10^{ - 5}}{\rm{\;T}}$

$5 \times {10^{ - 5}}{\rm{\;T}}$

JEE PHYSICS MCQ: MAGNETISM AND MATTER

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Application +4 | -1

The angle of dip at a place is

37^\circ

and the vertical component of the earth’s magnetic field is

6 \times {10^{ - 5}}\mathrm{\;T}.

The earth’s magnetic field at this place is

(\mathrm{tan\;}37^\circ = 3/4

)

View answer and explanation

Correct option: D

{10^{ - 4}}\;\mathrm{T}

Option D (

{10^{ - 4}}\;\mathrm{T}

) is correct because it is calculated using the given vertical component and the angle of dip.

Practice All Chapter MCQs →

{2^{5/4}}

X

Y

\mu_{\mathrm{d}}, \mu_{\mathrm{p}}

\mu_{\mathrm{f}}