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Electromagnetic Induction: Self and Mutual Inductance

An inductor consists of 500 turns of wire of resistance 6.0 Ω wound tightly and uniformly on a toroidal ring of an insulating magnetic material with relative permeability μ = 40. The material is linear so μ is independent of the magnetic field. The mean radius of the toroid is 15 cm and the cross sectional area is 1.0 cm2.

(there are 4 parts to this question (fulll quesion in attachment, the first is)

a)
Starting from Ampere's law, calculate the self inductance of the coil, assuming that the cross sectional area of the coil is sufficiently small that we can assume that the magnetic field has constant strength inside the coil.

Please can you fully show your working to help me understand. thank you.

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Question.

An inductor consists of 500 turns of wire of resistance 6.0 Ω wound tightly and uniformly on a toroidal ring of an insulating magnetic material with relative permeability μ = 40. The material is linear so μ is independent of the magnetic field. The mean radius of the toroid is 15 cm and the cross sectional area is 1.0 cm2.

a)
Starting from Ampere's law, calculate the self inductance of the coil, assuming that the cross sectional area of the coil is sufficiently small that we can assume that the magnetic field has constant strength inside the coil.

The magnetic field inside a long toroid of small cross section is same as in a long solenoid and can be derived using Ampere's law as bellow.

Let the toroid of mean radius r has N turns and carries a current I. Let magnetic field along the central line of the toroid is B. Considering an Amperian loop of radius r, the field at each point of the loop will be B and tangential to the loop.

According to Ampere's law the line integral of the ...

Solution Summary

A good question with four parts to learn Electromagnetic induction.
1. To derive and calculate self-inductance of a toroidal coil. 2 To calculate energy stored. 3 To calculate mutual-inductance. 4 To derive and calculate induced current.

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