Question

A fighter plane of length $20\, m$, wing span (distance from tip of one wing to the tip of the other wing) of $15\, m$ and height $5 \,m$ is flying towards east over Delhi. Its speed is $240 \, ms^{-1}$. 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. $V_B = 45 \, m V ; V_W = 120 \, mV$ with right side of pilot at higher voltage.
• B. $V_B = 45 \, mV ; V_W = 120 \, wV$ with left side of pilot at higher voltage
• C. $V_B = 40 \, mV , V_W = 135 mV$ with right side of pilot at high voltage
• D. $V_B = 40 \, mV ; V_W = 135 \, mV$ with left side of pilot at higher voltage

Answer 1

Answer: (B)

$V_{B}=B_{4}(5)(240)$
$B_{H}=B \cos \theta$
$B_{H}=\frac{5 \sqrt{5} \times 10^{-5}}{3}$
$B _{ V }=\frac{10}{3} \times 10^{-5} T$
$V _{ B }=\frac{5 \sqrt{5}}{3} \times 10^{-5} \times 5 \times 240$
$V_{B}=44.6 \,mV =45\, mV$
$V_{w}=B_{V} \ell V$
$=10^{-4} \times 1200$
$V_{\bigcup}=120\, mV$
(left side at fighter voltage)