Question

Query : A toroidal solenoid with an air core has an average radius of 15 cm, area of cross-section 12cm² and 1200 turns.
a. Obtain the self inductance of a toroid.
b. A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from 0 to 2 A in 0.05 s. Obtain the induced emf in the second coil.
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Answer 1

Given that,

• Radius = 15 cm

• Area of cross - section = 12 cm².

• No of turns = 1200.

The self inductance of toroid is given by :

⇒ I = uN²A/2πr

⇒ I = 2×10^-7(1200)²×12×10^-4/0.15

⇒ I = 0.000023 = 2.3 mH

∴ The self inductance of the toroid is 2.3 mH.

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Answer 2

Refer to above attachments for solution..

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