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A metallic square loop ABCD is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the figure. Find a) In which sides of the loop electric field is induced. b) Net emf induced in the loop. c) If one side .BC. is outside the field with remaining loop in the field and is being pulled out with a constant velocity then induced current in the loop. |
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Answer» Solution : a) The METALLIC SQUARE loop moves in its own plane with VELOCITY V. A uniform magnetic field is imposed perpendicular to the plane of the square loop. AD and BC are `bot`to the velocity as well as I to field applied. Hence ELECTRIC field is induced ACROSS the sides AD and BC only. b) As there is no change of flux through the entire coil net emf induced in the coil is zero. c) Induced current `i = e/R` Where R is the resistance of the coil. ` rArr i = (Blv)/( R)`(Only the side AD cuts the flux ) |
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