Please help - how do I magnetize water in test tube with electromagnetic field

AI Thread Summary
To magnetize water in a test tube effectively, it is important to understand that water itself is not ferromagnetic and will not retain a magnetic field. The discussion highlights that while wrapping a test tube with magnetic wire and applying a DC current can influence iron shavings, the water will not exhibit permanent magnetism. The effects observed may be due to the strong external magnetic field rather than the water itself. Additionally, the concept of the Curie Point and types of magnetism, such as paramagnetic and diamagnetic, are relevant to understanding this phenomenon. Ultimately, water cannot be magnetized in a way that allows it to retain or move iron shavings independently.
walela1
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Homework Statement


Trying to magnetize water in a test tube so that it will move or pick up iron shaving(s)


Homework Equations


none


The Attempt at a Solution


Wrapped test tube with magnetic wire (500 turns) wire is connected to DC current of 122 volts. It will pick up iron shavings with an iron rod in the test tube. It will not move or pick up shavings when I put only water in the test tube. How can I magnetize water in a test tube so that it will move or pick up iron shavings
 
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You might start to think about the various types of magnetism: paramagnetic, diamagnetic and ferromagnetic. You might also want to understand the definition of Curie Point in relation to ferromagnetism.

One point, if you are applying a strong magnetic field to the water, how will you know if the iron filings are moving in response to the water or the much larger applied magnetic field.

Hint: Water will not hold a magnetic field for long no matter how strong the applied field. When you switch off the applied magnetic field, the magnetic field generated by the water will decay to zero very quickly.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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