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JEE Advanced
Exam
JEE Advanced
Subject
Physics
Chapter
Magnetic Properties of Matter

JEE Advanced · Physics · Magnetic Properties of Matter

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

The angle of dip at a place is 37∘andtheverticalcomponentoftheearth′smagneticfieldis37^{{}^{{}^{\circ}}} and the vertical component of the earth's magnetic field is6 \times 10^{-5} \mathrm{~T}$. The earth's magnetic field at this place is \left(\tan 37^{{}^{{}^{\circ}}}=\frac{3}{4}\right)

Options

  1. A7×10−5 T7 \times 10^{-5} \mathrm{~T}
  2. B6×10−5 T6 \times 10^{-5} \mathrm{~T}
  3. C5×10−5 T5 \times 10^{-5} \mathrm{~T}
  4. D10−4 T10^{-4} \mathrm{~T}

Correct answer

D. 10−4 T10^{-4} \mathrm{~T}

Step-by-step solution

tan⁡θ=VH=34[∵tan⁡37$circ=34]\tan \theta=\frac{\mathrm{V}}{\mathrm{H}}=\frac{3}{4} \left[\because \tan 37^{\$circ}=\frac{3}{4}\right]
∴V=34H<br/>\therefore \mathrm{V}=\frac{3}{4} \mathrm{H}<br />\mathrm{V}=6 \times 10^{-5} \mathrm{~T}<br/><br />\mathrm{H}=\frac{4}{3} \times6×10−5 T=8×10−5 T6 \times 10^{-5} \mathrm{~T}=8 \times 10^{-5} \mathrm{~T}
∴Btotal =V2+$mathrmH2=(36+64)\therefore \mathrm{B}_{\text {total }}=\sqrt{\mathrm{V}^{2}+\$mathrmH^{2}}=\sqrt{(36+64)} \times 10^{-5}<br/><br />=10 \times 10^{-5}=10^{-4} \mathrm{~T} .$$
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