KCET · Physics · Magnetic Properties of Matter
A jet plane having a wing span of \( 25 \mathrm{~m} \) is travelling horizontally towards east with a speed of
\( 3600 \mathrm{~km} / \mathrm{h}^{-1} \).If the Earth's magnetic field at the location is \( 4 \times 10^{-4} \mathrm{~T} \) and the angle of dip is
\( 30^{\circ} \), then the potential difference between the ends of the wing is
- A \( 4 \mathrm{~V} \)
- B \( 5 \mathrm{~V} \)
- C O V
- D \( 2.5 \mathrm{~V} \)
Answer & Solution
Correct Answer
(B) \( 5 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
Given, wing span \(=25 \mathrm{~m}\); speed of jet plane \(=3600 \mathrm{kmh}^{-1}\)
Earth's magnetic field \(4 \times 10^{-4} \mathrm{~T} ;\) angle of dip \(=30^{\circ}\)
Now emf, \(e=B_{v} l v \rightarrow(1)\)
\(B_{v}=B \sin \theta=4 \times 10^{-4} \sin 30^{\circ}=4 \times 10^{-4} \times \frac{1}{2}=2 \times 10^{-4} \mathrm{~T}\)
\(v=3600 \mathrm{kmh}^{-1}=3600 \times \frac{5}{18}=1000 \mathrm{~ms}^{-1}\)
\(l=25 \mathrm{~m}\)
Substituting values in Eq. \((1)\), we get
\(e=2 \times 10^{-4} \times 1000 \times 25=5 \mathrm{~V}\)
Earth's magnetic field \(4 \times 10^{-4} \mathrm{~T} ;\) angle of dip \(=30^{\circ}\)
Now emf, \(e=B_{v} l v \rightarrow(1)\)
\(B_{v}=B \sin \theta=4 \times 10^{-4} \sin 30^{\circ}=4 \times 10^{-4} \times \frac{1}{2}=2 \times 10^{-4} \mathrm{~T}\)
\(v=3600 \mathrm{kmh}^{-1}=3600 \times \frac{5}{18}=1000 \mathrm{~ms}^{-1}\)
\(l=25 \mathrm{~m}\)
Substituting values in Eq. \((1)\), we get
\(e=2 \times 10^{-4} \times 1000 \times 25=5 \mathrm{~V}\)
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