Q131 / 180 · COMEDK 2014 · Physics
An aeroplane is flying horizontally with a velocity of $360 \mathrm{~km} \mathrm{~h}^{-1}$. The distance between the tips of the wings of the aeroplane is $50 \mathrm{~m}$. The vertical component of the earth's magnetic field is $4 \times 10^{-4} \mathrm{Wbm}^{-2}$. The induced emf is
- A$200 \mathrm{~V}$
- B$20 \mathrm{~V}$
- Correct answerC$2 \mathrm{~V}$
- D$0.2 \mathrm{~V}$
Solution
Given, velocity of aeroplane,
$$
\begin{aligned}
v &=360 \mathrm{kmh}^{-1} \\
&=360 \times \frac{5}{18} \mathrm{~ms}^{-1}=100 \mathrm{~ms}^{-1}
\end{aligned}
$$
Distance between the tips of wings,
$$
l=50 \mathrm{~m}
$$
Vertical component of earth's magnetic field,
$$
\begin{aligned}
&B_{V}=4 \times 10^{-4} \mathrm{Wbm}^{-2} \\
&\begin{aligned}
\therefore \text { Induced emf, } e &=B_{V} v l \\
&=4 \times 10^{-4} \times 100 \times 50=2 \mathrm{~V}
\end{aligned}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.