The angle of dip at a place is 37^ ^ and the vertical component of the earth's magnetic field is 6 × 10^ -5 ~T . The earth's magnetic field at this place is ( 37^ ^ =(3)/(4))

Correct answer: D. 10^ -4 T θ= V H =(3)/(4)[ 37^ ^ =(3)/(4)] V=(3)/(4) HV=6 × 10^ -5 TH=(4)/(3) × 6 × 10^ -5 T=8 × 10^ -5 T B_ total = V^ 2 +H^ 2 = (36+64) × 10^ -5 =10 × 10^ -5 =10^ -4 T.

BITSAT 2021

Exam: BITSAT
Year: 2021
Subject: Physics
Chapter: Magnetic Properties of Matter

BITSAT 2021 · Physics · Magnetic Properties of Matter

The angle of dip at a place is 37^ ^ and the vertical component of...

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

\left(\tan 37^{{}^{\circ}}=\frac{3}{4}\right)

Options

A $7 \times 10^{-5} \mathrm{ T}$
B $6 \times 10^{-5} \mathrm{ T}$
C $5 \times 10^{-5} \mathrm{ T}$
D $10^{-4} \mathrm{ T}$

Correct answer

D. $10^{-4} \mathrm{ T}$

Step-by-step solution

\tan \theta=\frac{\mathrm{V}}{\mathrm{H}}=\frac{3}{4}\left[\because \tan 37^{{}^{\circ}}=\frac{3}{4}\right]\therefore \mathrm{V}=\frac{3}{4} \mathrm{H}\mathrm{V}=6 \times 10^{-5} \mathrm{ T}\mathrm{H}=\frac{4}{3} \times 6 \times 10^{-5} \mathrm{ T}=8 \times 10^{-5} \mathrm{ T}\therefore \mathrm{B}_{\text{total }}=\sqrt{\mathrm{V}^{2}+\mathrm{H}^{2}}=\sqrt{(36+64)} \times 10^{-5}=10 \times 10^{-5}=10^{-4} \mathrm{ T}.

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