Question

A fighter plane of length $20m$, wing span (distance from tip of one wing to the tip of the other wing) of $15m$ and height $5m$ is flying towards east over Delhi. Its speed is $240m{s}^{-1}$. The earth’s magnetic field over Delhi is $5\times 10^{-5}T$ with the declination angle $\sim 0^{\circ }$ and dip of $\theta$ such that $\sin \theta =\frac{2}{3}$. If the voltage developed is $V_B$ between the lower and upper side of the plane and $V_W$ between the tips of the wings then $V_B$ and $V_W$ are close to :

• A. $V_B=45mV;V_W=120mV$ with right side of pilot at higher voltage.
• B. $V_B=45mV;V_W=120wV$ with left side of pilot at higher voltage
• C. $V_B=40mV,V_W=135mV$ with right side of pilot at high voltage
• D. $V_B=40mV;V_W=135mV$ with left side of pilot at higher voltage

Answer 1

Answer: (B)

$V_B=B_4(5)(240)$
$B_H=B\cos \theta$
$B_H=\frac{5\sqrt{5}\times 10^{-5}}{3}$
$B_V=\frac{10}{3}\times 10^{-5}T$
$V_B=\frac{5\sqrt{5}}{3}\times 10^{-5}\times 5\times 240$
$V_B=44.6mV=45mV$
$V_w=B_V\ell V$
$=10^{-4}\times 1200$
$V_{\bigcup }=120mV$
(left side at fighter voltage)