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Resistance of aluminum jumping ring

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Assume you have a solenoid with a diameter of 4.0 cm and 2900 turns per meter, with an impedance of 0.70. An iron core consisting of a bundle of wires is placed in the core. The core has an effective susceptibility of 10 (rather than 1000, for this is based on experimental results). An aluminum ring of a mass of 23 grams (1.4 cm length and inside diameter of 5.7cm) and thickness of 0.33 cm is placed around the protruding core. It operates at 120 V(rms), 60 Hz household ac.

If all of the flux from the solenoid passes through the ring, calculate that flux as a function of time. Find the induced emf from the changing flux.

Calculate the resistance of the ring, using a value of 3.3 x 10^(-8) ohm/m for the resistivity of the aluminum, and from it, find the current in the ring.

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The solution discusses the resistance of aluminum jumping ring.

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Solution:

Magnetic flux density B is produced by a solenoid wound, with n turns per unit length and current I passing through. It is given by: B = ?0?rnl (1)
This is where ?0 = permeability of free space and ?r = relative permeability of the core.

Furthermore, ?r = 1 + ?m = 1 + 10 = 11 (The susceptibility ?m of the core is given as 10)

The voltage applied to the solenoid is 120 V (rms), 60 Hz. Hence, the peak voltage is V0 = ?2 x 120 = 169.7 V. The applied voltage V = V0sin(2?f)t = 169.7sin377t. The magnitude of current I, through the solenoid, is ...

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