Questions About Magnetism Interfering With Electrical Circuits

[Bararontok]

It would be appreciated if formulas that would directly or indirectly allow the following quantities in question to be calculated are posted in this thread. The questions are listed below:

1.) How strong would a magnetic field have to be to block a given amount of electric power in a given electrical circuit?

2.) How much mass or volume would a permanent magnet need to generate a magnetic field of a particular strength?

3.) How much power and time would an electromagnet have to use to magnetize a magnetically coercive metal and enable the metal to generate a permanent magnetic field of a given strength or alternatively to demagnetize it?

4.) How do the dimensions of an inductor such as the number of turns, the diameter of the wire, the overall length of the wire and the diameter of the cylindrical cavity inside the inductor affect the amount of electricity the inductor will draw and the power of the electromagnetic field it will generate?

 

[Andrew Mason]

It would be appreciated if formulas that would directly or indirectly allow the following quantities in question to be calculated are posted in this thread. The questions are listed below:

1.) How strong would a magnetic field have to be to block a given amount of electric power in a given electrical circuit?

A constant magnetic field will not interfere with an electric current (although it can result in a force on moving charges or on a conductor carrying a current). The ability of a magnetic field to interfere with current flow in a circuit depends on the time rate of change of that magnetic field over the area enclosed by that circuit.

2.) How much mass or volume would a permanent magnet need to generate a magnetic field of a particular strength?

It does not depend on mass. I think it depends on the kind of material in the magnet (how many little magnetic dipoles it contains) and how aligned all the little magnetic dipoles are.

3.) How much power and time would an electromagnet have to use to magnetize a magnetically coercive metal and enable the metal to generate a permanent magnetic field of a given strength or alternatively to demagnetize it?

I think it would depend on the material and the forces opposing the alignment of all those little magnetic dipoles.

4.) How do the dimensions of an inductor such as the number of turns, the diameter of the wire, the overall length of the wire and the diameter of the cylindrical cavity inside the inductor affect the amount of electricity the inductor will draw and the power of the electromagnetic field it will generate?

If the coil's length is much longer than its diameter and uses a constant voltage source, the relationship between the magnetic field inside the core (B), number of turns (N), length of coil (L), permeability of the core ($\mu$) and current (I) approaches :

$$B = \mu NI/L$$

The amount of current it will draw after a brief start-up time is determined by Ohm's law: I = V/R where R is the resistance of the wire.

If an alternating voltage source is applied, it is more complicated. You have to take into account the inductive reactance which is a function of frequency (dI/dt) and the inductance (L in Henrys) as well as the resistance.

AM

 

[Bararontok]

A constant magnetic field will not interfere with an electric current (although it can result in a force on moving charges or on a conductor carrying a current). The ability of a magnetic field to interfere with current flow in a circuit depends on the time rate of change of that magnetic field over the area enclosed by that circuit.

The type of magnetic field being referred to is a dynamic magnetic field with circular magnetic field lines surrounding a permanent magnet and not an electrostatic field caused by a difference in the distribution of electric charge in a material.

It does not depend on mass. I think it depends on the kind of material in the magnet (how many little magnetic dipoles it contains) and how aligned all the little magnetic dipoles are.

Yes, the formula that takes magnetic energy density into account is also important because by multiplying the magnetic energy density by the mass, the total magnetic energy can be taken. But the question is, how does magnetic energy equate to magnetic field strength?

I think it would depend on the material and the forces opposing the alignment of all those little magnetic dipoles.

If the coil's length is much longer than its diameter and uses a constant voltage source, the relationship between the magnetic field inside the core (B), number of turns (N), length of coil (L), permeability of the core ($\mu$) and current (I) approaches :

$$B = \mu NI/L$$

The amount of current it will draw after a brief start-up time is determined by Ohm's law: I = V/R where R is the resistance of the wire.

If an alternating voltage source is applied, it is more complicated. You have to take into account the inductive reactance which is a function of frequency (dI/dt) and the inductance (L in Henrys) as well as the resistance.

AM

The formula relating electric power to magnetic field strength is needed, because this formula only refers to the core but what about the inductor itself? How much power would the inductor need to generate a magnetic field that will magnetize a metallic core of high magnetic coercivity so that it will maintain a permanent magnetic field even after the inductor is turned off?

 

[Andrew Mason]

The type of magnetic field being referred to is a dynamic magnetic field with circular magnetic field lines surrounding a permanent magnet and not an electrostatic field caused by a difference in the distribution of electric charge in a material.

I am not clear on whether you are referring to a time-dependent magnetic field (eg one that surrounds a conductor carrying a time-dependent current) or a constant field of a permanent magnet. Only a time-dependent magnetic field (ie not a constant field) will interfere with an electric current.

Yes, the formula that takes magnetic energy density into account is also important because by multiplying the magnetic energy density by the mass, the total magnetic energy can be taken. But the question is, how does magnetic energy equate to magnetic field strength?

The energy stored in the magnetic field of an inductor (inductance = L) is:

$$E = \frac{1}{2}LI^2$$

The energy density of that magnetic field is derived http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html" . I am not sure if that helps you.

The formula relating electric power to magnetic field strength is needed, because this formula only refers to the core but what about the inductor itself? How much power would the inductor need to generate a magnetic field that will magnetize a metallic core of high magnetic coercivity so that it will maintain a permanent magnetic field even after the inductor is turned off?

That is a complicated thing to analyse that depends on a number of factors that you have not given us. The energy needed to magnetise a material is likely much greater than the energy that is stored in the magnetic field after the material has been permanently magnetised - ie. there is heat lost in getting all those little magnetic dipoles turned all the same way.

AM

 

[Bararontok]

Okay, those questions have been answered. Now, since an inductor stores energy via a magnetic field, is it possible for the inductor to retain a time dependent magnetic field even if the power source is cut after the inductor is fully charged and the inductor is left in an open circuit configuration?

 

[Naty1]

nope...and it's probably not useful to think of an inductor as "fully charged" which would apply to a capacitor. Likely you mean that a steady state magnetic field configuration has been established. The magnetic field energy is deried from the current passing thru it.

In other words you can't get a shock from an inductor as you can from a capacitor after power is shut off...

 

[Bararontok]

The inductor generates a reverse current when the power drops to zero due to the collapse of the magnetic field, but of course, even if the circuit is open this would still happen right?

 

[Antiphon]

The inductor generates a reverse current when the power drops to zero due to the collapse of the magnetic field, but of course, even if the circuit is open this would still happen right?

It generates a reverse voltage, not a reverse current. Yes, it happen when, not if the circuit is opened.

The right way to say it "charging a capacitor" and "magnetizing and inductor".

 

[Antiphon]

nope...and it's probably not useful to think of an inductor as "fully charged" which would apply to a capacitor. Likely you mean that a steady state magnetic field configuration has been established. The magnetic field energy is deried from the current passing thru it.

In other words you can't get a shock from an inductor as you can from a capacitor after power is shut off...

Actually you can but it's not a common thing. The mathematical dual of a capacitor holding a charge is an inductor circulating a current.

If you get a current flowing through an (ideal) inductor and short the terminals, the current will flow indefinately. If you then touch the leads and break the circuit, you most certainly will get a shock.

 

[Bararontok]

So because the reverse voltage is only generated if the inductor is closed circuited, then does it maintain the magnetic field if it is magnetized and then the circuit is opened?

 

[Naty1]

of course not. You'd have inductors sitting around in labs all magnetized up!