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Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of solenoid 2 is 0.0320 $\mathrm{Wb}$ (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is $2.54 \mathrm{A},$ what is the average flux through each turn of solenoid 1 ?

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a,$[1.96 \mathrm{H}]$b. $=7.11 \times 10^{-3} \mathrm{Wb}$

Physics 102 Electricity and Magnetism

Chapter 21

Electromagnetic Induction

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Inductance

Alternating Current

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

Lectures

03:27

Electromagnetic induction is the production of an electromotive force (emf) across a conductor due to its dynamic interaction with a magnetic field. Michael Faraday is generally credited with the discovery of electromagnetic induction in 1831.

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In physics, a magnetic field is a vector field that describes the magnetic influence of electric currents and magnetic materials. The magnetic field at any given point is specified by both a direction and a magnitude (or strength); as such it is a vector field. The term is used for two distinct but closely related fields denoted by the symbols B and H, where H is measured in units of amperes per meter (usually in the cgs system of units) and B is measured in teslas (SI units).

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Two solenoids $A$ and $B$ …

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Two solenoids A and B, spa…

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Two solenoids $A$ and $B$,…

04:43

this problem. We have two coils that are wrapped around the same terrible form. We're told that one of the coils had 700 turns and a current of 6.52 ants running through it. So this drawn, you can think of that being the green coil and for the other coil, the red coil that has 400 turns. And we're told that for this current running through the first coil, it experiences an average flux through one of these turns of 10.32 leaders. Our first task is to find the mutual induct in CE. So the form of this that will want to use is our equation for and that flux running through this second coil divided by our current running through the first. And we have all of these. So its program in 400 turns 0.3 to river and divided by 6.5 grams. If we put this thing, we get 1.9 600. Her part B. We want to find what the average flux is through one of the turns in coil. One if we have a current of 2.54 amps running through the second quarter. But this we're gonna want to exploit the fact that the mutual induct Ince's the same for both coils. So we'll use the same equation from part, eh? But now in the numerator will have been that flux through the first coil, divided by the current in the second coil. We solve this for the average flux, running through one of the coils. One of the turns in the first coil have Am I to over anyone? And now, since we know what the Mutual induct Ince's from part A, we can plug it in 1.96 Henry. We're told that this is 2.54 amps running through it, and we have 700 turns. So we plug all this in get 7.11 times 10 to the 3 10 to the negative three Weaver or 7.11 Really, Weaver

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