Why Does a Slinky Show a Sine Wave When Measuring Magnetic Fields?

AI Thread Summary
The discussion revolves around a physics lab measuring the magnetic field inside a slinky with an electric current, which produces a sine wave graph. This sine wave is attributed to the alternating current (AC) used in the experiment, as the mains power supply generates an oscillating current. The magnetic field inside the coil can be approximated using the formula B = μ₀ I N, where I represents the current and N is the number of turns. In an AC circuit, the current can be expressed as I₀ = (V₀/|z|)sin(ωt + φ), with φ determined by the circuit's impedance. Clarification on these equations is offered for those unfamiliar with the concepts.
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I'm trying to do this lab in physics where we are using sensors to find the magnetic field within a slinky with a current flowing through. Our graph(of time vs.magnetic field) keeps coming out as a sine wave. Is the wave equation for an electromagnetic wave applicable to find the magnitude of the magnetic field? My teacher and I are both a little bit confused, so some guidance would be appreciated. Thanks!
 
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did you plug the slinky into the wall ? the mains power supply is alternating current.

the b field inside the coil is approximated by

B= \mu _o I N
where N is the number of turns and I is the current. in an AC circuit I would be

I_o = \frac{V_o}{|z|}sin(\omega t +\phi) where \phi is determined by
<br /> \phi = tan(\frac{\chi_L}{R})
and
\chi _L = \omega L

if you don't understand of this say so and i will try to explain
 
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