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 lying 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^o}$ 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

  • [JEE MAIN 2016]
  • A

    $V_B = 40\, mV$; $V_W = 135\,mV$ with left side of pilot at higher voltage

  • B

    $V_B = 45\,mV$; $V_W = 120\, mV$ with right side of pilot at higher voltage

  • C

    $V_B= 40\, mV$; $V_W = 135\,mV$ with right side of pilot at higher voltage

  • D

    $V_B = 45\, mV$; $V_W = 120\, mV$ with left side of pilot at higher voltage

Similar Questions

Two short bar magnets of magnetic moments $m$ each are arranged at the opposite corners of a square of side $d$ such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is

Two normal uniform magnetic field contain a magnetic needle making an angle $60°$  with $F$. Then the ratio of $\frac{F}{H}$ is

A dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is $40°$. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of $30°$ with the magnetic meridian. In this position the needle will dip by an angle

The lines joining the places of the same horizontal intensity are known as

The direction of the null points is on the equatorial line of a bar magnet, when the north pole of the magnet is pointing