A toroidal solenoid with air core of has an average radius of 15 cm, area of cross section 12 cm2 and has 1200 turns. Calculate self-inductance of the toroid. Assume field to be uniform across the cross section of toroid.

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Given that r=15 cm =0.15 mA=12 cm2=12×10-4 m2N=1200 turnsThe self inductance of toroid is given by L=μoN2A2πrL=2×10-7×12002×12×10-40.15L=0.0023 HL=2.3 mH

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