Question

A metal aeroplane having a distance of 50 meter between the edges of its wings is flying horizontally with a speed of 360 km/hour. At the place of flight, the earth’s total magnetic field is 4.0 × 10^–5weber/meter² and the angle of dip is 30^o. The induced potential difference between the edges of its wings will be

• A. 0.1 V
• B. 0.01 V
• C. 1 V
• D. 10 V

Answer 1

Answer: (A)

0.1 V

Answer 2

The flying of the aeroplane is shown in the adjoining figure. Its wings are cutting flux-lines due to the vertical component of earth’s magnetic field. So, a potential difference V (say) in induced between the edges of its wings.

If the earth’s total magnetic field be B and the angle of dip θ, then the vertical component of earth’s magnetic field is B_V = B sinθ = (4.0 × 10^–5) × sin 30^o = 2.0 × 10^–5 wb/metre².

We know that when a conductor of length l meter moves with velocity v meter/second perpendicular to a magnetic field of B weber/meter², then the potential difference induced in the conductor is given by $V = B_{V}vlvolt$

Here l = 50 meter, v = 360 km/hour

$= \frac{360 \times 1000}{60 \times 60} = 100meter/second$

and B = B_V = 2.0 × 10^–5 weber/meter² ;

$V = (2.0 \times 10^{- 5}) \times 100 \times 50 = 0.1volt$