Understanding Power in Solenoids for MRI

[physicsnoobie]

so far I've been studying that P= work done (energy change) / time = I x V
and its all been good... until i came across P= I squared x R (i know how this is derived) when studying solenoids in Magnetic Resonance Imaging and this was referred to as the power loss (energy dissipation over time)

1. i found this somewhat confusing.. since it suggests that the process: electrical energy -> heat energy (losses), is the only energy change process that is going on in a solenoid

i thought about it and so far I am guessing that maybe this is true in solenoids but not in normal electric circuits because electrical energy -> heat energy (losses) + light energy (bulbs)

anyway, this was in the context of why copper is not used in the solenoid of MRI.. because the large combined resistance would result in large power losses.. instead superconductors with zero/negligible resistance is used

2. then again, does that mean that power of the solenoid in MRI = zero/infinitely small because the resistance is negligible? (following P = I squared x R)

anyway, thanks to anyone who shares some light on this.. hopefully in simple terms.. I am quite a newbie in physics :(

 

[Gear300]

The power when considering P = I^2*R is explicitly noted as the rate at which energy is supplied to a resistance. The power of the solenoid, I'm guessing, would be the rate at which energy is supplied to the magnetic field in the solenoid. If I remember correctly, the energy of the magnetic field of a solenoid may be given as U = (1/2)LI^2; so if the current is constant, the power is 0W. In the case of non-steady state circuits or AC circuits, the current does change.

 

[astrokreng]

I can provide some clarification on the concept of power in solenoids for MRI. First, it is important to understand that power is the rate at which energy is transferred or converted. In the context of solenoids, power refers to the rate at which electrical energy is converted into other forms, such as heat or mechanical energy.

The equation P = I x V is the general formula for power, where I represents the current flowing through the solenoid and V represents the voltage applied. This equation is valid for all electrical systems, including solenoids used in MRI. However, in the case of MRI, there is an additional factor to consider - the resistance of the solenoid.

Resistance is a measure of how much a material opposes the flow of electrical current. In normal electric circuits, as you mentioned, resistance leads to the conversion of electrical energy into heat and light energy. However, in the case of MRI solenoids, the resistance is kept as low as possible to minimize power losses. This is because the primary purpose of the solenoid in MRI is to create a strong magnetic field, and any energy lost due to resistance would result in a weaker magnetic field.

So, in the context of MRI solenoids, the equation P = I squared x R is used to calculate power losses. This equation takes into account the resistance of the solenoid, which is why it is referred to as power loss. It is important to note that this equation does not mean that the only energy change process in a solenoid is electrical energy being converted to heat energy. There are other energy changes happening, such as the conversion of electrical energy into magnetic energy to create the magnetic field.

Now, to address your second question, the power of the solenoid in MRI is not zero or infinitely small. While the resistance of superconducting materials used in MRI solenoids may be negligible, there are still power losses due to other factors, such as imperfections in the material or external magnetic fields. However, these losses are significantly lower compared to using materials with higher resistance, which is why superconductors are used in MRI solenoids.

I hope this helps to clarify the concept of power in solenoids for MRI. It is a complex topic, but understanding the different factors at play can help in further understanding the functioning of MRI machines. Keep learning and exploring!