Question

Question: An aeroplane with its wings spread 10 m is flying with speed 180 kph in

horizontal direction. The total intensity of earth’s field is 4 2.5 10− × Tesla and angle of dip is

60 . ° Then find emf induced between the tips of the plane wings.

Options:

(a) 108 mV

(b) 54 mV

(c) 216 mV

(d) 140 mV​

Answer 1

CORRECT 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/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings.

ANSWER:

• The emf induced between the tips of the plane wings = 108 mV.

GIVEN:

• Length of wings = 10 m.

• Speed of aeroplane = 180 kmph.

• Earth's magnetic field = 2.5 × 10^{-4} Wb/m²

TO FIND:

• The emf induced between the tips of the plane wings.

EXPLANATION:

$\bigstar \red{ \boxed{ \bold{ \large{e = \vec{B} \ \vec{l} \ \vec{v} }}}} \\ \\ \\ \tt\pink{ Here \ \vec{B} \ \perp \ \vec{l} \ \perp \ \vec{v} }\\ \\ \\ \tt \green{Imagine\ the\ velocity\ is \ in \ x-direction.}\\ \\ \\ \tt \blue{Then\ we\ need\ vertical\ component} \\ \tt \blue{of\ Earth's \ magnetic\ field \ which} \\ \tt \blue{will\ be\ in\ y-direction.} \\ \\ \\ \texttt{\orange{We already know that the wings are}}\\ \texttt{\orange{perpendicular to the velocity }}\\ \texttt{\orange{and so to the vertical component of }}\\ \texttt{\orange{Earth's magnetic field. So it will }} \\ \texttt{\orange{be in z-direction.}} \\ \\ \\ \tt \vec{B} \ \vec{l} \ \vec{v} = B \ sin {60}^{ \circ} \times 10 \times 180 \times \dfrac{5}{18} \\ \\ \\ \tt \vec{B} \ \vec{l} \ \vec{v} = 2.5 \times {10}^{ - 4} \times \frac{ \sqrt{3} }{2} \times 10 \times 50 \\ \\ \\ \tt\vec{B} \ \vec{l} \ \vec{v} = 2.5 \times 1.732 \times 250 \times {10}^{ - 4} \\ \\ \\ \\ \tt\vec{B} \ \vec{l} \ \vec{v} =625 \times 1.732 \times {10}^{ - 4} \\ \\ \\ \\ \tt\vec{B} \ \vec{l} \ \vec{v} =1082.5 \times {10}^{ - 4} \implies108.2 \ mV$

Hence the emf induced between the tips of the plane wings = 108 mV.

Answer 2

$\maltese \: \: { \underline{ \underline{ \bf{ \red{Question:}}}}}$

• An aeroplane with its wings spread 10 m is flying with speed 180 kph in horizontal direction. The total intensity of earth’s field is 2.5× 10^-4Tesla and angle of dip is 60° . Then find emf induced between the tips of the plane wings.

$\maltese \: \: { \underline{ \underline{ \bf{ \red{Answer:}}}}}$

$\looparrowright \bf \pink{option \: (a) \: 108 \: mV}$

$\maltese \: { \underline{ \underline{ \red{ \bf{Given :}}}}}$

$\bullet \tt \: B \: = 2.5 \times {10}^{ - 4} \: \: T \\ \\ \bull \tt \: δ = 60 \degree$

$\maltese \: \: { \underline{ \underline{ \red{ \bf{To \: \: Find:}}}}}$

• Then find emf induced between the tips of the plane wings.

$\maltese \: \: { \underline{ \underline{ \red{ \bf{Solution:}}}}}$

we know that,

$\underline {\huge{ \boxed{ \rm{ \blue{B_v = B \: sin \: δ}}}}} \: \bigstar$

$\qquad \hookrightarrow \rm \: B_v = 2.5 \times {10}^{ - 4} \: sin \: 60\degree$

$\qquad \: \tt \: sin \: 60 \degree = \frac{ \sqrt{3} }{2}$

$\qquad \hookrightarrow \rm \: B_v = \frac{2.5 \sqrt{3} }{2} \times {10}^{ - 4}$

Then ,

$\qquad \hookrightarrow \rm l \: = 10 \: m$

$\qquad \hookrightarrow \rm \: V = 180km/ h$

$\qquad \hookrightarrow \rm \: V = \frac{180 \times 5}{18} \: \: m/ s$

$\qquad \hookrightarrow \rm \: V = 50 \: m/ s$

$\underline{ \huge{ \boxed{ \rm{ \blue{|E|=B_v \: l \: v}}}}} \bigstar$

$\qquad \looparrowright \rm \: |E| = \frac{2.5 \sqrt{3} }{2} \times {10}^{ - 4} \times 10 \times 50$

$\qquad \looparrowright \rm \: |E| = 1082 \times {10}^{ - 4}V$

$\qquad \looparrowright \rm \: |E| = 108.2\times {10}^{ - 3}V$

$\qquad \looparrowright {\rm{\orange {|E| = 108 \: mV}}}$

•°• The emf induced between the tips of the plane wings is 108 mV.