ExamBro
ExamBro
KCET · Physics · Magnetic Properties of Matter

An aircraft with a wingspan of \( 40 \mathrm{~m} \) flies with a speed of \( 1080 \mathrm{~km} \mathrm{~h}^{-1} \) in the eastward
direction at a constant altitude in the northern hemisphere, where the vertical component of the
earth's magnetic field \( 1.75 \times 10^{-5} \mathrm{~T} \). Then the emf developed between the tips of the wings is

  1. A \( 0.5 \mathrm{~V} \)
  2. B \( 0.34 \mathrm{~V} \)
  3. C \( 0.21 \mathrm{~V} \)
  4. D \( 2.1 \mathrm{~V} \)
Verified Solution

Answer & Solution

Correct Answer

(C) \( 0.21 \mathrm{~V} \)

Step-by-step Solution

Detailed explanation

Given, length of wing, \(l=40 \mathrm{~m}\); speed, \(v=1080 \mathrm{kmh}^{-1}\)
\(=1080 \times \frac{5}{18} \mathrm{~ms}^{-1}\)
Earth's magnetic field,\(B=1.75 \times 10^{-5} \mathrm{~T}\)
Then emf
\(E=B l v=\left(1.75 \times 10^{-5}\right) \times 40 \times 1080 \times \frac{5}{18}\)
\(=21000 \times 10^{-5}=0.21 \mathrm{~V}\)
Therefore, the emf developed between the tips of the wing is \(0.21 \mathrm{~V}\)