A fighter plane of length 20 m, wing span (distane 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. It speed is `240 ms^1` . The earth’s magnetic field over Delhi is `5 xx 10^5` T with the declination angle `~ 0^0` and dip of `theta` such that `sin=(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 :

Options

A. `V_(B) = 45 mV, V_(w) = 120 mV` with right side of pilot at higher voltage
B. `V_(B) = 45 mV, V_(w) = 120 mV` with left 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) = 40 mV, V_(w) = 135 mV` with left side of pilot at higher voltage

Answer

B

Explanation

Similarly as last problem, only angle of dip will change so
`sin theta =(2)/(3)`(given)
`v= 240 m//s` `B_(e)= 5 xx 10^(-5)T`
`B_(V)= B_(e)sintheta= 5 xx 10^(-5) xx (2)/(3)`
`(1)/(3) xx 10^(-4) T`
`V_(w)` induced emf while cutting the vertical component of Earth.s magnetic field by wings of plane
`epsilon = B_(v)vl`
`= (1)/(3) xx 10^(-4) xx 240 xx 15`
`= 12 xx 10^(-2)`
`epsilon = V_(w)= 0.12V= 120mV`
Now, `V_(B)` induced emf while cutting the horizontal component of Earth.s magnetic field by lower and upper side of the plane
`V_(b)= epsilon_(1)= B_(H)vl_(1)`
`B_(0)costheta vl_(1)`
`5 xx 10^(-5)(sqrt(1- sin^(2)theta)) xx 240 xx l_(1)`
`5 xx 10^(-5)(sqrt(1- (4)/(9))) xx 240 xx 5`
= 45mV
By Fleming.s left hand rule, Left side of the pilot is at higher voltage.