MHT CET · Physics · Magnetic Properties of Matter
If the angle of dip at places \(\mathrm{A}\) and \(\mathrm{B}\) are \(30^{\circ}\) and \(45^{\circ}\) respectively, the ratio of horizontal component of earth's magnetic field at \(A\) to that at \(B\) will be
\([\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin \frac{\pi}{6}=\frac{1}{2},\) \(\quad \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}]\)
- A \(\sqrt{2}: 1\)
- B \(1: \sqrt{2}\)
- C \(\sqrt{2}: \sqrt{3}\)
- D \(\sqrt{3}: \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3}: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
(B)
The horizontal component of earth's magnetic field is given by,\(H=R \cos \delta\) where, \(\delta\) is the angle of dip so \(\mathrm{H}_{1}=\mathrm{R} \cos 30^{\circ}\)
and \(\mathrm{H}_{2}=\mathrm{R} \cos 45^{\circ}\)
\(\therefore \frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R} \cos 30^{\circ}}{\mathrm{R} \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\)
The horizontal component of earth's magnetic field is given by,\(H=R \cos \delta\) where, \(\delta\) is the angle of dip so \(\mathrm{H}_{1}=\mathrm{R} \cos 30^{\circ}\)
and \(\mathrm{H}_{2}=\mathrm{R} \cos 45^{\circ}\)
\(\therefore \frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R} \cos 30^{\circ}}{\mathrm{R} \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\)
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