KCET · Physics · Magnetic Properties of Matter
The horizontal component of Earth's magnetic field at a place is \(3 \times 10^{-5} \mathrm{~T}\). If the dip at that place is \(45^{\circ}\), the resultant magnetic field at that place is
- A \(3 \times 10^{-5} \mathrm{~T}\)
- B \(\frac{3}{\sqrt{2}} \times 10^{-5} \mathrm{~T}\)
- C \(3 / 2 \sqrt{3} \times 10^{-5} \mathrm{~T}\)
- D \(3 \sqrt{2} \times 10^{-5} \mathrm{~T}\)
Answer & Solution
Correct Answer
(D) \(3 \sqrt{2} \times 10^{-5} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
Given,
\(B_H =3 \times 10^{-5} \mathrm{~T}\)
\(\delta =45\)
From the diagram
\(\cos \delta=\frac{B_H}{B}\)
\(\begin{aligned} \Rightarrow B & =\frac{B_H}{\cos \delta}=\frac{3 \times 10^{-5}}{\cos 45^{\circ}} \\ & =3 \sqrt{2} \times 10^{-5} \mathrm{~T}\end{aligned}\)
\(B_H =3 \times 10^{-5} \mathrm{~T}\)
\(\delta =45\)
From the diagram
\(\cos \delta=\frac{B_H}{B}\)
\(\begin{aligned} \Rightarrow B & =\frac{B_H}{\cos \delta}=\frac{3 \times 10^{-5}}{\cos 45^{\circ}} \\ & =3 \sqrt{2} \times 10^{-5} \mathrm{~T}\end{aligned}\)
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