JEE Mains · Physics · STD 12 - Magnetism and matter
- JEE Mains 2016
- Hard
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(V_B = 45\, mV\); \(V_W = 120\, mV\) with left side of pilot at higher voltage
Step-by-step Solution
Detailed explanation
\(\mathrm{V}_{\mathrm{B}}=\mathrm{VB}_{\mathrm{H}} l=240 \times 5 \times 10^{-5} \cos (\theta) \times 5\)
\(=44.7\, \mathrm{mV}\)
By right hand rule, the charge moves to the left of pilot.
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