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# Toroidal Solenoid

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A toroidal solenoid has 550 turns, cross- sectional area 6.7 square cm, and a mean radius of 4.8 cm.

A) Calculate the coil's self-inductance.

B) If the current decreases uniformly from 5A to 2A in 3ms, calculate the self induced emf in the coil.

C) The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a? Explain.

https://brainmass.com/physics/emf/toroidal-solenoid-234855

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Please refer to the attachment for a complete solution with diagrams.

A) Self inductance of a solenoid is given by : L = Î¼0N2A/L where N = Total number of turns = 550, A = Area of cross section = 6.7 cm2 = 6.7x10-4 m2, L = Length = 4.8 cm = 0.048 m. Substituting values, we get :

L = 4Î x10-7x(550)2x6.7x10-4/0.048 = 5.3x10-3 H or 5.3 mH

B) Self induced emf magnitude = e = L(dI/dt) where dI = change in current = 3A, dt = time in which the change in current takes place = 3x10-3 sec. Hence,

e = 5.3x10-3x3/3x10-3 = 5.3 Volts

C) As per Lenz's law, direction of the induced emf is such as to oppose the cause of the change in the flux.

(Please see attached file for diagram)

Let the solenoid be connected to a battery of variable emf as shown in the fig.. Initially 5A current flows through the solenoid. The battery emf is now reduced gradually so as to bring the current to 2A in 3 ms. The flux passing through the solenoid also reduces correspondingly. The emf induced in the coil will be such as to oppose the cause of the reduction in the flux i.e. reduction in the battery emf. Hence, polarity of the induced emf will be to add to the battery emf

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