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JEE Mains · Physics · STD 12 -6. Electromagnetic induction

An aeroplane, with its wings spread \(10\, m ,\) is flying at a speed of \(180 \,km / h\) in a horizontal direction. The total intensity of earth's field at that part is \(2.5 \times 10^{-4}\, Wb / m ^{2}\) and the angle of dip is \(60^{\circ}\). The emf induced between the tips of the plane wings will be ...... \(mV\)

  1. A \(108.25\)
  2. B \(54.125\)
  3. C \(88.37\)
  4. D \(62.50\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(108.25\)

Step-by-step Solution

Detailed explanation

\(\in=[\vec{B} \vec{V} \vec{L}]=B V L \sin \theta\) \(=\left(2.5 \times 10^{-4}\, T \right)\left(180 \times \frac{5}{18}\, m / s \right)(10 \,m ) \sin 60^{\circ}\) \(=108.25 \times 10^{-3}\, V\)
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