(C) Given: Wingspan $l = 40 \text{ m}$,speed $v = 1080 \text{ km h}^{-1}$.Convert speed to $SI$ units: $v = 1080 \times \frac{5}{18} \text{ m s}^{-1}

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(C) Given: Wingspan $l = 40 \text{ m}$,speed $v = 1080 \text{ km h}^{-1}$.Convert speed to $SI$ units: $v = 1080 \times \frac{5}{18} \text{ m s}^{-1} = 300 \text{ m s}^{-1}$.Vertical component of Earth's magnetic field $B = 1.75 \times 10^{-5} \text{ T}$.The motional emf induced across the wingspan is given by $E = B l v$.Substituting the values: $E = (1.75 \times 10^{-5} \text{ T}) \times (40 \text{ m}) \times (300 \text{ m s}^{-1})$.$E = 1.75 \times 10^{-5} \times 12000 = 1.75 \times 0.12 = 0.21 \text{ V}$.Thus,the emf developed between the tips of the wings is $0.21 \text{ V}$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

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An aircraft with a wingspan of $40 \text{ m}$ flies with a speed of $1080 \text{ km h}^{-1}$ in the eastward direction at a constant altitude in the northern hemisphere,where the vertical component of the earth's magnetic field is $1.75 \times 10^{-5} \text{ T}$. The emf developed between the tips of the wings is: (in $\text{ V}$)

KCET 2015 All KCET

A
$0.5$
B
$0.34$
C
$0.21$
D
$2.1$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

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$A$ conducting rod $PQ$ of length $5\,m$ oriented as shown in the figure is moving with velocity $(2\,m/s)\hat{i}$ without any rotation in a uniform magnetic field $(3\hat{j} + 4\hat{k})\,T$. The $Emf$ induced in the rod is.....$V$.

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