CET Karnataka Medical
CET - Karnataka Medical Solved Paper-2015
question_answer
An aircraft with a wingspan of 40 m flies with a speed of 1080 km/hr in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of the earths magnetic field \[1.75\times {{10}^{-5}}T\]. Then the emf developed between the tips of the wings is
A)\[0.34V\]
B)\[2.1V\]
C)\[0.5V\]
D)\[0.21V\]
Correct Answer:
D
Solution :
: The emf developed between the tips of the wings is \[\varepsilon ={{B}_{V}}{{l}_{V}}\] Here, \[{{B}_{V}}=1.75\times {{10}^{-5}}T,\,l=40m\] \[v=1080km/hr=1080\times \frac{5}{18}m\,{{s}^{-1}}\] \[=300\,m\,{{s}^{-1}}\] \[\therefore \] \[\varepsilon =(1.75\times {{10}^{-5}}T)(40m)(300\,m\,{{s}^{-1}})\] \[=0.21V\]