Questions involving electromagnetic induction at your level are usually simple and interesting. Here are the two questions which appeared in Kerala Engineering Entrance 2008 question paper:
(1) If the self inductance of 500 turn coil is 125 mH, then the self inductance of similar coil of 800 turns is
(a) 48.8 mH
(b) 200 mH
(c) 187.5 mH
(d) 320 mH
(e) 78.1 mH
Self inductance is the total magnetic flux linked with the coil when unit current flows through the coil. Since the total flux and the magnetic field are directly proportional to the number of turns in the coil, the self inductance is directly proportional to the square of the number of turns in the coil. Therefore we have
5002/8002 = 125/L where L is the self inductance of the 800 turn coil.
This yields L = 320 mH.
(2) The flux linked with a circuit is given by Φ= t3 + 3t – 7. The graph between time (x–axis) and induced emf (y–axis) will be a
(a) straight line through the origin
(b) straight line with positive intercept
(c) straight line with negative intercept
(d) parabola through the origin
(e) parabola not through the origin
The induced emf, V = – dΦ/dt = – 3t2 – 3
If V is plotted against t, a parabola which does not pass through the origin will be obtained. So the correct option is (e).
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