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Unformatted text preview: EP106 General Physics II Page 1/10 11. INDUCTANCE 111 Self Induction If two coils near each other, a current i in one coil will set up a flux & B through the second coil. If this flux changes in time an induced emf will appear in the second coil according to Faradays Law. This is called self induction and the emf produced is called self induced emf (or back emf ). For the second coil: dt N d B ) (  = here the product N & B is the number of flux linkages ( N being number of turn). For a given coil (if no magnetic materials such as iron nearby) this quantity is proportional to to the current i . Li N B = or i N L B = where L is a proportionality constant  called the inductance of the coil. This is analogous to the defining the capacitance: V Q C = The inductance depends on the geometry of the device. Hence: dt di L dt N d B =  = ) ( or dt di L /  = The SI unit of L is Henry (H) & 1 H = 1 V s/A We can find the direction of a selfinduced emf from Lenz Law. Assume that he instantaneous currents below are the same in ach case self induced emf L is directed so as to oppose the change. the current is decreasing the current is increasing EP106 General Physics II Page 2/10 EXAMPLE 1 Find the inductance of a uniformly wound solenoid having N turns and length & . Assume that & is much longer than the radius of the windings and that the core of the solenoid is air. SOLUTION We can assume that the magnetic field inside a solenoid is uniform. Thus: i N ni B & = = where & / N n = is the number of turns per unit length. The flux is: i NA BA B & = = where A is the crosssectional area of the solenoid. Finally, the inductance is: & A N i N L B 2 = = or if we use & n N = & A n L 2 = Ques : What would happen to the inductance if a ferromagnetic material were placed inside the solenoid? Ans : The inductance would increase. For a given current, the magnetic flux is now much greater because of the increase in the field originating from the magnetization of the ferromagnetic material. For example, if the material has a magnetic permeability of 500 , the inductance would increase by a factor of 500. EXAMPLE 2 (a) Calculate the inductance of an aircore solenoid containing 300 turns if the length of the solenoid is 25.0 cm and its crosssectional area is 4 cm 2 . (b) Calculate the selfinduced emf in the solenoid if the current through it is decreasing at the rate of 50 A/s....
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This note was uploaded on 02/29/2012 for the course PHYS 227 taught by Professor Rabe during the Fall '08 term at Rutgers.
 Fall '08
 RABE
 Physics, Current, Inductance

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