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 × 10-4 Wb / m 2 and the angle of dip is 60° . The emf induced between the tips of the plane wings will be :-

Question

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 × 10-4 Wb / m 2 and the angle of dip is 60° . The emf induced between the tips of the plane wings will be :- (A) 108.25 mV (B) 54.125 mV (C) 88.37 mV (D) 62.50 mV

Tardigrade Admin · Accepted Answer

∈=[ vec B vec V vec L ]= BVL sin θ =(2.5 × 10-4 T )(180 × (5/18) m / s )(10 m ) sin 60° =108.25 × 10-3 V

$108.25 mV$

$54.125 mV$

$88.37 mV$

$62.50 mV$

$\in= [B V L] = B V L sin θ$
$= (2.5 \times 1 0^{- 4} T) (180 \times \frac{5}{18} m / s) (10 m) sin 6 0^{\circ}$
$= 108.25 \times 1 0^{- 3} V$

Q. An aeroplane, with its wings spread 10m, is flying at a speed of 180km/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 :-

108.25mV

54.125mV

88.37mV

62.50mV

∈=[BVL]=BVLsinθ =(2.5×10−4T)(180×185​m/s)(10m)sin60∘ =108.25×10−3V

$108.25 mV$

$54.125 mV$

$88.37 mV$

$62.50 mV$

$\in= [B V L] = B V L sin θ$
$= (2.5 \times 1 0^{- 4} T) (180 \times \frac{5}{18} m / s) (10 m) sin 6 0^{\circ}$
$= 108.25 \times 1 0^{- 3} V$