Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.10
NCERT Solutions for Class 12 Physics Chapter 6 Question 10
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Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 x 10-4 T
Angle of dip, δ = 30°
Vertical component of Earth’s magnetic field,
Bv = B sin δ = 5 x 10_4sin 30°
= 2.5 x 10-4 T
Voltage difference between the ends of the wing can be calculated as: e = (Bv) x l x v
= 2.5 x 10⁻⁴x 25 x 500 = 3.125 V
Hence, the voltage difference developed between the ends of the wings is 3.125 V.