To learn about mutual inductance from an example of a long solenoid with two windings. To illustrate the calculation of mutual inductance it is helpful to consider the specific example of two solenoids that are wound on a common cylinder. We will take the cylinder to have radius p and length L. Assume that the solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length: § B(7) - dỉ = µ-encl (Figure 1) We will consider the field that arises from solenoid 1, which has n₁ turns per unit length. The magnetic field due to solenoid 1 passes (entirely, in this case) through solenoid 2, which has №₂ turns (total turns not turns per length). Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction, E() dl=м(t).

To learn about mutual inductance from an example of a long solenoid with two windings. To illustrate the calculation of mutual inductance it is helpful to consider the specific example of two solenoids that are wound on a common cylinder. We will take the cylinder to have radius p and length L. Assume that the solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length: § B(7) - dỉ = µ-encl (Figure 1) We will consider the field that arises from solenoid 1, which has n₁ turns per unit length. The magnetic field due to solenoid 1 passes (entirely, in this case) through solenoid 2, which has №₂ turns (total turns not turns per length). Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction, E() dl=м(t).

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter13: Electromagnetic Induction

Section: Chapter Questions

Problem 13.9CYU: Check Your Understanding A long solenoid of cross-section area 5.0 cm2is wound with 25 turns of wire...

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