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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 5×10−5 T with the declination angle ~0^{\circ} and dip of θ such that   \sin \Theta = \frac{2}{3}    If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to :

  • Option 1)

     VB = 45 mV ;  VW = 120 mV with right side of pilot at higher voltage

  • Option 2)

    VB = 45 mV ; VW = 120 mV with left side of pilot at higher voltage

  • Option 3)

    VB = 40 mV ; VW = 135 mV with right side of pilot at higher voltage

  • Option 4)

    VB = 40 mV ; VW = 135 mV with left side of pilot at higher voltage

 

Answers (1)

best_answer

As we have learned

Motional EMF -

\varepsilon = Blv
 

- wherein

B\rightarrow magnetic field

l\rightarrow length

u\rightarrow velocity of u perpendicular to uniform magnetic field.

 

 

V_B = B_H\: \: 5 \times 240 \\ B_H = B \cos \theta \\ B_H = \frac{5 \sqrt 5 \times 10^{-5}}{3}

B_V = 10 / 3 \times 10^{-5}T

V_B = \frac{5 \sqrt 5 }{3 } \times 10^{-5}\times 5 \times 240 = 45 m_v

V_w = B_v l_v = 10 ^{ -4 }\times 1200\\=120 mv

 

 

 

 

 

 

 

 

 


Option 1)

 VB = 45 mV ;  VW = 120 mV with right side of pilot at higher voltage

Option 2)

VB = 45 mV ; VW = 120 mV with left side of pilot at higher voltage

Option 3)

VB = 40 mV ; VW = 135 mV with right side of pilot at higher voltage

Option 4)

VB = 40 mV ; VW = 135 mV with left side of pilot at higher voltage

Posted by

Avinash

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