Search any question & find its solution
Question:
Answered & Verified by Expert
A jet plane is travelling towards west at a speed of 1800 $\mathrm{km} / \mathrm{h}$. What is the voltage difference developed between the ends of the wing having a span of $25 \mathrm{~m}$, if the Earth's magnetic field at the location has a magnitude of $5 \times 10^{-4} \mathrm{~T}$ and the dip angle is $30^{\circ}$.
Solution:
1008 Upvotes
Verified Answer
Given: vielocity v $=1800 \mathrm{~km} / \mathrm{h}$.
$$
=\frac{1800 \times 1000}{60 \times 60}=500 \mathrm{~m} / \mathrm{s}
$$
length $\ell=25 \mathrm{~m}$
earth's field $\mathrm{B}=5 \times 10^{-4} \mathrm{~T}$
angle of dip $\delta=30^{\circ}$
To find: voltage developed or emf induced $\mathrm{e}=$ ?
Formula used: $e=B \ell v$
Since plane travels horizontally it is acted upon by vertical component of earth's field
$$
\begin{aligned}
&\mathrm{V}=\mathrm{B} \sin \delta=5 \times 10^{-4} \times \sin 30^{\circ}=5 \times 10^{-4} \times \frac{1}{2} \\
&=2.5 \times 10^{-4} \mathrm{~T}
\end{aligned}
$$
Thusinduced emfe $=\mathrm{B} \ell \mathrm{v}=2.5 \times 10^{-4} \times 25 \times 500=3.125 \mathrm{~V}$
$$
=\frac{1800 \times 1000}{60 \times 60}=500 \mathrm{~m} / \mathrm{s}
$$
length $\ell=25 \mathrm{~m}$
earth's field $\mathrm{B}=5 \times 10^{-4} \mathrm{~T}$
angle of dip $\delta=30^{\circ}$
To find: voltage developed or emf induced $\mathrm{e}=$ ?
Formula used: $e=B \ell v$
Since plane travels horizontally it is acted upon by vertical component of earth's field
$$
\begin{aligned}
&\mathrm{V}=\mathrm{B} \sin \delta=5 \times 10^{-4} \times \sin 30^{\circ}=5 \times 10^{-4} \times \frac{1}{2} \\
&=2.5 \times 10^{-4} \mathrm{~T}
\end{aligned}
$$
Thusinduced emfe $=\mathrm{B} \ell \mathrm{v}=2.5 \times 10^{-4} \times 25 \times 500=3.125 \mathrm{~V}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.