Question
In a certain place, the horizontal component of magnetic field is $\frac{1}{\sqrt{3}}$ times the vertical component. The angle of dip at this place is(a) zero(b) $\pi / 3$(c) $\pi / 2$(d) $\pi / 6$
Step 1
It is given by the formula: \[ \tan(\theta) = \frac{B_v}{B_h} \] where \(B_v\) is the vertical component of the magnetic field and \(B_h\) is the horizontal component of the magnetic field. Show more…
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Key Concepts
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The horizontal component of earth’s magnetic field at a place is √3 times the vertical component. The angle of dip at that place is (i) π/6 (ii) π/3 (iii) π/4 (iv) 0
At a certain place, the horizontal component of the earth's magnetic field is $B_{0}$ and the angle of dip is $45^{\circ}$. The total intensity of the field at that place will be (a) $B_{0}$ (b) $\sqrt{2} B_{0}$ (c) $2 B_{0}$ (d) $B_{0}^{1}$
Magnetostatics
Round 1
The real angle of dip, if a magnet is suspended at an angle of $30^{\circ}$ to the magnetic meridian and the dip needle makes an angle of $45^{\circ}$ with horizontal, is : (a) $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ (b) $\tan ^{-1}(\sqrt{3})$ (c) $\tan ^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}}\right)$ (d) $\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
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