Question

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 500 turns, while solenoid 2 has 300 turns. When the current in solenoid 1 is 3.21A, the average flux through each turn of solenoid 2 is 0.0120Wb. When the current in solenoid 2 is 1.23A, find the average flux through each turn of solenoid 1.

          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 500 turns, while solenoid 2 has 300 turns. When the current in
solenoid 1 is 3.21A, the average flux through each turn of solenoid 2 is
0.0120Wb. When the current in solenoid 2 is 1.23A, find the average flux
through each turn of solenoid 1.
        
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Added by Smith T.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 13th Edition
Chapter 30
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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 500 turns, while solenoid 2 has 300 turns. When the current in solenoid 1 is 3.21A, the average flux through each turn of solenoid 2 is 0.0120Wb. When the current in solenoid 2 is 1.23A, find the average flux through each turn of solenoid 1.
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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 500 turns, while solenoid 2 has 300 turns. When the current in solenoid 1 is 3.21A, the average flux through each turn of solenoid 2 is 0.0120Wb. When the current in solenoid 2 is 1.23A, find the average flux through each turn of solenoid 1.

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Transcript

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00:01 Here in this given problem, number of turns in the first toroid means n1, that is 500, and current passing through it, that is 3 .21 amper.
00:28 And the number of turns in the second toroid, that is, n2, is equal to 300, and current passing through it, that current passing from.
00:40 Through it that is 1 .23 ampere flux linked through the second toroid average flux linked through per turn of the second toroid due to the current in the first one 5 to 1 this is given at 0 .0 120 weber and average flux linked through each turn of the first to the current in the second one that is missing we have to find so using an expression for the flux that is b a product of the magnetic field with the area when the magnetic field lines are perpendicular to the area and an expression for the magnetic field in the toroid that is mu not n i or we can say mu not total turns capital n divided by length because this small n is the number of turns per unit length and length of toroid means its circumference 2 pi r as both the toroid are same so their area and radius will be same hence we can say phi 2 1 that will be given by mu not n1 i 1 divided by 2 pi r and multiplied by a as a and r are same and phi linked through the first coil due to the current in the second one, phi 1 -2 that will be mu -0 n2 i2 divided by 2 by r multiplied by a...
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