CBSE Class 12 · 2018 · Electromagnetic InductionLong Answer

An aeroplane is flying horizontally from west to east with a velocity of

900 km/hour

. Calculate the potential difference developed between the ends of its wings having a span of

20 m

. The horizontal component of the Earth's magnetic field is

5 \times 1 0^{- 4} T

and the angle of dip is

3 0^{\circ}

Answer: $ε \approx 1.44 V$ .

Solution

Convert the speed:

v = 900 km/h = 900 \times \frac{1000}{3600} = 250 m/s

, and the wing span is

l = 20 m

.
The horizontal wings, moving horizontally, cut the vertical component of Earth's field, so the motional emf is

ε = B_{V} l v

.
The vertical component is

B_{V} = B_{H} tan δ = (5 \times 1 0^{- 4}) tan 3 0^{\circ} = \frac{5 \times 1 0 ^{- 4}}{3} \approx 2.89 \times 1 0^{- 4} T .

Therefore

ε = B_{V} l v = (2.89 \times 1 0^{- 4}) (20) (250) \approx 1.44 V .

ε \approx 1.44 V

CBSE Class 12 Board · 2018 · Subjective