Consider the wing of 20m span of a jet plane which is travelling west at a speed of 1,600kmph . The Earth's magnetic field at a location has a magnitude of 0.5× 10- 3T and the dip angle 30° . Find the voltage difference developed between the ends of the wing at this location:

Question

Consider the wing of 20m span of a jet plane which is travelling west at a speed of 1,600kmph . The Earth's magnetic field at a location has a magnitude of 0.5× 10- 3T and the dip angle 30° . Find the voltage difference developed between the ends of the wing at this location: (A) 4.1V (B) 2.2V (C) 3.2V (D) 3.8V

Tardigrade Admin · Accepted Answer

Here, v=1600 km h- 1=1600× (5/18)=444m s- 1 l=20 m, B=5× 10- 4T and δ=30o As ϵ =Blv=Vlv (Where V is the vertical component of earthâs field) =(.Bsin δ.)lv =5× 10- 4sin 30o × 20 × 444 = 2.2 V

$4.1 V$

$2.2 V$

$3.2 V$

$3.8 V$

Here, $v = 1600 km h^{- 1} = 1600 \times \frac{5}{18} = 444 m s^{- 1}$
$l = 20 m, B = 5 \times 1 0^{- 4} T$ and $δ = 3 0^{o}$
As $ϵ = Bl v = V l v$
(Where $V$ is the vertical component of earth’s field)
$= (B s in δ) l v$
$= 5 \times 1 0^{- 4} s in 3 0^{o} \times 20 \times 444 = 2.2 V$

Q. Consider the wing of 20m span of a jet plane which is travelling west at a speed of 1,600kmph . The Earth's magnetic field at a location has a magnitude of 0.5×10−3T and the dip angle 30∘ . Find the voltage difference developed between the ends of the wing at this location:

4.1V

2.2V

3.2V

3.8V

Here, v=1600kmh−1=1600×185​=444ms−1 l=20m,B=5×10−4T and δ=30o As ϵ=Blv=Vlv (Where V is the vertical component of earth’s field) =(Bsinδ)lv =5×10−4sin30o×20×444=2.2V

$4.1 V$

$2.2 V$

$3.2 V$

$3.8 V$

Here, $v = 1600 km h^{- 1} = 1600 \times \frac{5}{18} = 444 m s^{- 1}$
$l = 20 m, B = 5 \times 1 0^{- 4} T$ and $δ = 3 0^{o}$
As $ϵ = Bl v = V l v$
(Where $V$ is the vertical component of earth’s field)
$= (B s in δ) l v$
$= 5 \times 1 0^{- 4} s in 3 0^{o} \times 20 \times 444 = 2.2 V$