An aeroplane, with its wings spread 10m, 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 × 10-4Wb/m2 and the angle of dip is 60°. The emf induced between the tips of the plane wings will be…………..

Bv = B sin 60°-> B v = 2 . 5 × 1 0 − 4 × 3 2 <!- [if gte mso 9]><xml> </xml><! [endif]-> E m f = B v × v × l = 2 . 5 × 1 0 − 4 × 3 2 × 1 8 0 × 5 1 8 × 1 = 1 0 8 . 2 5 × 1 0 − 3 v o l t s <!- [if gte mso 9]><xml> </xml><! [endif]->

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Electromagnetic Induction Physics Ncert Solutions Class 12th Physics Class 12th

An aeroplane, with its wings spread 10m, 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 × 10^-⁴Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be…………..

Option 1 -

88.37mV

Option 2 -

54.125mV

Option 3 -

62.50mV

Option 4 -

108.25mV

3 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 4

Detailed Solution:

B_v = B sin 60°

-> $B_{v} = 2.5 \times 1 0^{- 4} \times \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} \times v \times l = 2.5 \times 1 0^{- 4} \times \frac{\sqrt[]{3}}{2} \times 180 \times \frac{5}{18} \times 1 = 108.25 \times 1 0^{- 3} v o l t s$

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An aeroplane, with its wings spread 10m, 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 × 10^-⁴Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be…………..

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