JEE Mains · Physics · STD 12 - 5. Magnetism and matter
The vertical component of the earth's magnetic field is \(6 \times 10^{-5} T\) at any place where the angle of dip is \(37^{\circ}\). The earth's resultant magnetic field at that place will be \(\left(\right.\) Given \(\left.\tan 37^{\circ}=\frac{3}{4}\right)\)
- A \(8 \times 10^{-5}\,T\)
- B \(6 \times 10^{-5}\,T\)
- C \(5 \times 10^{-4}\,T\)
- D \(1 \times 10^{-4}\,T\)
Answer & Solution
Correct Answer
(D) \(1 \times 10^{-4}\,T\)
Step-by-step Solution
Detailed explanation
\(\left(\delta=37^{\circ}\right)\) \(B _{ V }= B \sin \delta\) \(6 \times 10^{-5}= B \frac{3}{5}\) \(B =10 \times 10^{-5}\,T\) \(=10^{-4}\,T\)
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