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 ...... \(mV\)
- A \(108.25\)
- B \(54.125\)
- C \(88.37\)
- D \(62.50\)
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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