KCET · Physics · Magnetic Properties of Matter
At certain place, the horizontal component of earth's magnetic field is \( 3.0 \mathrm{G} \) and the angle dip
at that place is \( 30^{\circ} \). The magnetic field of earth at that location
- A O.5 G
- B \( 5.1 \mathrm{G} \)
- C \( 3.5 \mathrm{G} \)
- D \( 6.0 \mathrm{G} \)
Answer & Solution
Correct Answer
(C) \( 3.5 \mathrm{G} \)
Step-by-step Solution
Detailed explanation
Given, horizontal component of Earth's magnetic field \(=3.0 \mathrm{G} ;\) angle of dip \(=30^{\circ}\)
Now, horizontal component = B \(\cos\) (angle of dip)
\(\Rightarrow 3.0 G=B \cos 30^{\circ}\)
\(\Rightarrow B=\frac{3.0}{\cos 30^{\circ}}=\frac{3.0 \times 2}{\sqrt{3}}\)
\(\Rightarrow B=2 \sqrt{3}=3.464 \sim 3.5 \mathrm{G}\)
Therefore, magnetic field of Earth at that location \(=3.5 \mathrm{G}\)
Now, horizontal component = B \(\cos\) (angle of dip)
\(\Rightarrow 3.0 G=B \cos 30^{\circ}\)
\(\Rightarrow B=\frac{3.0}{\cos 30^{\circ}}=\frac{3.0 \times 2}{\sqrt{3}}\)
\(\Rightarrow B=2 \sqrt{3}=3.464 \sim 3.5 \mathrm{G}\)
Therefore, magnetic field of Earth at that location \(=3.5 \mathrm{G}\)
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