Validating a Modified Biot-Savart Law

[HAgdn]

Homework Statement

Need help if this equation that I thought of would indeed provide right values. I do not have a Tesla meter to compare my values with.

What I am trying to do is solving for the magnetic flux density of produced by a wire with AC voltage. So, I used ohms low (to use voltage as a variable) to calculate for the forwarding and reversing current

Where:
B is Magnetic Flux Density
Micro symbol is permeability
V_peak is the peak voltage of AC current
R is wire resistance (of the wire producing the magnetic field)
r is the distance from the wire producing the field

Relevant Equations

B(t)= μ[(V_peak (sin⁡ωt ))/R]/2πr

I tried validating the equation by somehow 'solving' only with the involved units.
Magnetic Flux Density – Tesla (T) – Weber (Wb) per square meter (Wb/m2)

 

[kuruman]

The fact that you have AC should not prevent you from replacing $I$ in Biot-Savart with $I_{peak}\sin(\omega t)$. Biot-Savart is applicable for all currents.

 

[HAgdn]

Is there any way to 'move' ω outside the sine?

 

[kuruman]

Is there any way to 'move' ω outside the sine?

Not in the Biot-Savart expression. The product ($\omega t$) is an angle in radians, strictly speaking a dimensionless quantity. $\omega$ can "move" outside either by taking a derivative or doing an integral involving $\sin(\omega t)$. Why are you asking? Does this problem involve Faraday's Law and magnetic flux due to an infinite wire carrying a time-varying current? If so, please provide the entire statement of the problem as given to you. Also, why do you think you need a Tesla meter? Is your question related to an experiment?

 

[HAgdn]

Sir, thank you for answering my post.

My question is indeed related to an experiment.
- That would be calculating the magnetic flux density (Tesla) produced by AC currents. I would like to compare my calculation results (using the equation) with that of data read from a Tesla meter.

I am looking for a way to pull out the ω from inside the sine as I would do an integral involving the sin(ωt), since ω (the given frequency of an AC) is constant, only t would remain.

 

[kuruman]

Sir, thank you for answering my post.

My question is indeed related to an experiment.
- That would be calculating the magnetic flux density (Tesla) produced by AC currents. I would like to compare my calculation results (using the equation) with that of data read from a Tesla meter.

I am looking for a way to pull out the ω from inside the sine as I would do an integral involving the sin(ωt), since ω (the given frequency of an AC) is constant, only t would remain.

If you need additional help, I suggest that you describe your experiment and provide diagrams of the setup, if possible. Be as specific as you can. Note that if your goal is to use Biot-Savart to find B (Tesla), you will have to do an integral over space, not over time, so there is no $\omega$ that is "pulled out." This means that the field B is proportional to the current I at all times. You have not explained why you believe you need to pull $\omega$ out.

 

[rude man]

Homework Statement:: Need help if this equation that I thought of would indeed provide right values. I do not have a Tesla meter to compare my values with.

What I am trying to do is solving for the magnetic flux density of produced by a wire with AC voltage. So, I used ohms low (to use voltage as a variable) to calculate for the forwarding and reversing current

Where:
B is Magnetic Flux Density
Micro symbol is permeability
V_peak is the peak voltage of AC current
R is wire resistance (of the wire producing the magnetic field)
r is the distance from the wire producing the field
Relevant Equations:: B(t)= μ[(V_peak (sin⁡ωt ))/R]/2πr

I tried validating the equation by somehow 'solving' only with the involved units.
Magnetic Flux Density – Tesla (T) – Weber (Wb) per square meter (Wb/m2)
View attachment 256130

B = $\mu I/2\pi r$ I understand, but the 3rd term simplifies to HA so A must = $\mu~$??

 

[HAgdn]

Sir, what do you mean by the third term simplifies to HA?

μ is the permeability and is with the unit, Henry per meter. It is also expressed in kilogram meter square per second square ampere square.

.

I tried doing the whole equation again and I seem to be getting AH/m^2 instead.

 

[rude man]

You're right, I made a silly mistake.
Like @kuruman I don't see why you question your "relevant equation". I also wonder why you wrote all those dimensional equivalents.

My knowledge of a B meter is limited to the Hall effect. The Hall probe measurs B as a linear function of voltage so there is no need to either differentiate or integrate the sine function.

Of course you are aware that Ampere's law, while universally correct, is limited in usefulness to very long wires.

 

[HAgdn]

Sir, I was questioning my relevant equation because it is involved in my research. I am having doubts as to if such an equation does represent reality, of course, it may not be accurate but, well, just close.

oh, they are called dimensional equivalents? I did those to check if the output value is indeed magnetic flux density. After all, the equation is composed of few laws.

 

[HAgdn]

If you need additional help, I suggest that you describe your experiment and provide diagrams of the setup, if possible. Be as specific as you can. Note that if your goal is to use Biot-Savart to find B (Tesla), you will have to do an integral over space, not over time, so there is no $\omega$ that is "pulled out." This means that the field B is proportional to the current I at all times. You have not explained why you believe you need to pull $\omega$ out.

Sir, $\omega$ is the frequency of the AC source. Therefore, it is constant.

 

[rude man]

Sir, I was questioning my relevant equation because it is involved in my research. I am having doubts as to if such an equation does represent reality, of course, it may not be accurate but, well, just close.

It is close if r << length of wire.

oh, they are called dimensional equivalents? I did those to check if the output value is indeed magnetic flux density. After all, the equation is composed of few laws.

2 laws: Ampere's and Ohm's: $i = \frac {2\pi r B} {\mu}$ and i = V/R. If you stick to rationalized MKS system you can't go wrong!

 

[kuruman]

Sir, I was questioning my relevant equation because it is involved in my research. I am having doubts as to if such an equation does represent reality, of course, it may not be accurate but, well, just close.

The equation $B=\dfrac{\mu_0~I}{2\pi r}$ gives the magnetic field $B$ (conventional units Tesla) at distance $r$ from a wire carrying current $I$. The wire is assumed to be "infinitely" long. Now anything physical like a wire cannot have a mathematically infinite length. For a finite wire of length $2L$, the equation is valid when $r~<<L$, i.e. the closer you are to the wire the better the approximation. The better the approximation the closer the equation is (in your words) to reality. Since you have not told us what this is all about, I cannot be more specific.

Sir, $\omega$ is the frequency of the AC source. Therefore, it is constant.

I agree, $\omega$ is constant. You still have not explained why you believe it is important "to pull $\omega$ out" of the sine argument. We are going around in circles here and you are not helping us help you.

 

[HAgdn]

I know one thing - that I know no thing. Sir is indeed right. I am confused that I do not know what to say.

Sir, what I know is that what I am trying to do is to create a mathematical model that would represent the magnetic flux density produced by a wire running with an AC source.

And it is known that magnetic flux density is calculated with the B formula B=μI/2πr [Equation 1].

Since the magnetic flux density source is AC, its current behavior must be taken into consideration.
so, I = I_peak(sin(ω *t)) [Equation 2]
This makes I_peak as the draw current or simply, the current through a the wire as the B source, while ω is the frequency, in my case, 60Hz. t is also considered as AC current is sinusoidal that changes over time.

With this the magnetic flux density is indeed changing, therefore can possibly induce emf on nearby conductive materials if the B flux density is strong.

What I am doubting is that is it really just valid to replace I in Equation 1 with the I in Equation 2?

To remove my doubts, I tried doing the dimensional equivalents to see if the unit of the output value is indeed magnetic flux density.

And Sir, thank you for your time.

 

[HAgdn]

The equation $B=\dfrac{\mu_0~I}{2\pi r}$ gives the magnetic field $B$ (conventional units Tesla) at distance $r$ from a wire carrying current $I$. The wire is assumed to be "infinitely" long. Now anything physical like a wire cannot have a mathematically infinite length. For a finite wire of length $2L$, the equation is valid when $r~<<L$, i.e. the closer you are to the wire the better the approximation. The better the approximation the closer the equation is (in your words) to reality. Since you have not told us what this is all about, I cannot be more specific.I agree, $\omega$ is constant. You still have not explained why you believe it is important "to pull $\omega$ out" of the sine argument. We are going around in circles here and you are not helping us help you.

Sir, I ditched the idea of pulling out ω from inside the sine. I had that idea of pulling it out since in integration, constants are somewhat separated from the changing variables.

 

[haruspex]

Sir, I ditched the idea of pulling out ω from inside the sine. I had that idea of pulling it out since in integration, constants are somewhat separated from the changing variables.

Integration is a linear process, so multiplicative constants can be pulled outside: ∫Af(x).dx=A∫f(x).dx. But your ω is inside a nonlinear function, so that cannot be pulled out.

 

[kuruman]

I know one thing - that I know no thing. Sir is indeed right. I am confused that I do not know what to say.

Sir, what I know is that what I am trying to do is to create a mathematical model that would represent the magnetic flux density produced by a wire running with an AC source.

And it is known that magnetic flux density is calculated with the B formula B=μI/2πr [Equation 1].

Since the magnetic flux density source is AC, its current behavior must be taken into consideration.
so, I = I_peak(sin(ω *t)) [Equation 2]
This makes I_peak as the draw current or simply, the current through a the wire as the B source, while ω is the frequency, in my case, 60Hz. t is also considered as AC current is sinusoidal that changes over time.

With this the magnetic flux density is indeed changing, therefore can possibly induce emf on nearby conductive materials if the B flux density is strong.

What I am doubting is that is it really just valid to replace I in Equation 1 with the I in Equation 2?

To remove my doubts, I tried doing the dimensional equivalents to see if the unit of the output value is indeed magnetic flux density.

And Sir, thank you for your time.

OK, if you accept that the right hand side of $B=\dfrac{\mu_0~I}{2\pi r}$ has the correct dimensions when the current $I$ is constant in time, then you must accept that when you replace this steady current with time-varying current $I_{peak}\sin(\omega t)$, the right hand side will still have correct dimensions. That's because $\sin(\omega t)$ is has no dimensions like $2$ and $\pi$ in the expression. What carries the dimension of Amperes is $I_{peak}.$

If you have doubts whether you should use the expression for your research, they should stem from whether the point(s) where you need to know the field is (are) far enough from the ends of the wire. In that case it might be more appropriate to use the expression for a finite segment of wire carrying current $I_{peak}\sin(\omega t)$ instead of $I$. See here for several examples of how to do this.