If the vertical component of the earth's magnetic field is 0.45 G at a location, and angle of dip is…
- 0.26 G
- 0.52 G
- 0.3 G
- 0.7 G
Solution

$\begin{aligned} & \mathrm{B}_{\mathrm{v}}=0.45 \mathrm{G}, \delta=60^{\circ} \\ & \therefore \quad \mathrm{B}_{\mathrm{v}}=\mathrm{b} \sin \delta \\ & \Rightarrow \mathrm{B}=\frac{\mathrm{B}_{\mathrm{v}}}{\sin \delta}=\frac{0.45}{\sin 60^{\circ}}=\frac{0.9}{\sqrt{3}}=0.52 \mathrm{G}\end{aligned}$
Asked in: AP EAMCET 2024 (21 May Shift 1)
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