Solving for x in Circular Current Equation

In summary: I just cancel x out and getIn summary, if you substitute the -983nV for the 983nV in the problem, you get the same answer.
  • #1
Jaccobtw
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Homework Statement
A very large loop of metal wire with radius 1m is driven with a linearly increasing current at a rate of 200A/s. A very small metal wire loop with radius 5cm is positioned a small distance away with its center on the same axis (the loops are coaxial). The small loop experiences an induced emf of 983nV. What is the separation of the loops in m? Note that a subtraction step in the solution makes it sensitive to significant figures. Keep at least four figures in your calculation.
Relevant Equations
$$\int E \cdot ds = \varepsilon = -\frac{d}{dt}\int B \cdot dA$$
$$B = \frac{\mu_0 I R^{2}}{2(x^{2} + R^{2})^{3/2}}$$
The second equation gives the magnetic field at a point away from the center of a circular current. If we multiply this by the the area we get a function for the magnetic flux. We have an increasing current which induces an increasing magnetic field. Now just solve for x in the second equation to get the distance the second loop is away from the first loop. But I don't think you can just plug in a changing current in for I. Where have I gone wrong? Thank you
 
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  • #2
Jaccobtw said:
Relevant Equations:: $$\int E \cdot ds = \varepsilon = -\frac{d}{dt}\int B \cdot dA$$ $$B = \frac{\mu_0 I R^{2}}{2(x^{2} + R^{2})^{3/2}}$$

The second equation gives the magnetic field at a point away from the center of a circular current. If we multiply this by the the area we get a function for the magnetic flux. We have an increasing current which induces an increasing magnetic field. Now just solve for x in the second equation to get the distance the second loop is away from the first loop. But I don't think you can just plug in a changing current in for I.

You can't solve for the value of ##x## using the second equation since you don't know the value ##B## or ##I##.
However, you are given the value of the induced emf. Can you use the first equation to solve for the value of ##x##?
 
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  • #3
You should work towards calculating the expression for the EMF on the small loop. It will turn out that it is a function of x (and ##\frac{dI}{dt}## and that's where the rate of change of current comes in). You wrote that you multiply B by the area to get the flux, that's correct, but what you should do to the flux expression to get the expression for the EMF?
 
  • #4
Delta2 said:
You should work towards calculating the expression for the EMF on the small loop. It will turn out that it is a function of x (and ##\frac{dI}{dt}## and that's where the rate of change of current comes in). You wrote that you multiply B by the area to get the flux, that's correct, but what you should do to the flux expression to get the expression for the EMF?
if you integrate the changing current with respect to time, you'll find that I = 200t. the problem is now that I have two variables for the magnetic field - t and x. How do I take the integral of B with respect to both those variables IF that's what I should do?
 
  • #5
Jaccobtw said:
if you integrate the changing current with respect to time, you'll find that I = 200t. the problem is now that I have two variables for the magnetic field - t and x. How do I take the integral of B with respect to both those variables IF that's what I should do?
You want to keep x as unknown cause that's what the problem is asking for. The time t will be simplified when you take the time derivative of the flux. We assume that the loop is small enough so that the B-field is homogeneous inside it. No need to do spatial integration to find the flux, a simple multiplication of the B-field at the center with the area of the small loop should suffice.
 
  • #6
Delta2 said:
You want to keep x as unknown cause that's what the problem is asking for. The time t will be simplified when you take the time derivative of the flux. We assume that the loop is small enough so that the B-field is homogeneous inside it. No need to do spatial integration to find the flux, a simple multiplication of the B-field at the center with the area of the small loop should suffice.
The problem is when I try to isolate x on one side I keep getting a negative inside a square root.
 
  • #7
Jaccobtw said:
The problem is when I try to isolate x on one side I keep getting a negative inside a square root.
Hmmm, do you get the correct answer if you assume that the induced EMF is -983nV instead of the 983nV that the problem states?
 
  • #8
Delta2 said:
Hmmm, do you get the correct answer if you assume that the induced EMF is -983nV instead of the 983nV that the problem states?
$$\varepsilon = -\frac{d}{dt} BA$$ 1
$$\frac{\varepsilon}{A} = -\frac{dB}{dt}$$ 2
$$\int\frac{\varepsilon dt}{A} = -\int dB = -B$$ 3
$$\frac{\varepsilon t}{A} = -\frac{\mu_0 200t R^{2}}{2(R^{2} + x^{2})^{3/2}}$$ 4
$$\varepsilon = -\frac{\mu_0 200 R^{2} A}{2(R^{2} + x^{2})^{3/2}}$$ 5, t cancels
$$(R^{2} + x^{2})^{3/2} = -\frac{\mu_0 200 R^{2} A}{2\varepsilon }$$ 6
After this I take both sides to the 2/3 power. Then because R has a value of 1, I subtract 1 from both sides then take the sqaure root of both sides. Because 1 is bigger than what it is subtracting, the value ends up being negative on the right side inside a square root. Not sure what I'm doing wrong
 
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  • #9
I think your approach is fundamentally flawed. When you write ##\varepsilon = -\frac{d}{dt} BA##, you take ##A## to be the area of the big loop. That is inappropriate. It is the changing magnetic flux at the position of the smaller loop that generates the emf. So in the above equation ##B## is the magnetic field at the location of the smaller loop and ##A## must be the area of the smaller loop. As @Delta2 already noted, you may assume that the magnetic field is uniform over the area of the loop and has its on-axis value.

Also note that the negative sign in front of the rate of change of flux is there because of Lenz's law. It indicates that the induced emf always opposes the rate of change of magnetic flux. If you apply Faraday's law formally with vectors and use the right hand rule to link the directed normal to the loop with the circulation of the line integral, you should have no problem with negative signs under radicals. If you are lazy like me, you can just take the absolute value and ignore the negative sign.
 
  • #10
My mistake. I wrote the permeability of free space as ##4 \times 10^{-7}## instead of ##4\pi \times 10^ {-7}##. All of that leads to the correct answer. Thanks alot
 
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FAQ: Solving for x in Circular Current Equation

What is the Circular Current Equation?

The Circular Current Equation is a mathematical formula used to calculate the strength of a circular magnetic field created by an electric current flowing through a circular loop.

Why is Solving for x in Circular Current Equation important?

Solving for x in the Circular Current Equation allows us to determine the magnitude of the magnetic field at a specific point in space, which is crucial for understanding the behavior of electrical currents and their effects on objects.

What are the variables in the Circular Current Equation?

The variables in the Circular Current Equation are the magnetic field strength (B), the current (I), the distance from the center of the loop (r), and the permeability of free space (μ0).

How do you solve for x in the Circular Current Equation?

To solve for x in the Circular Current Equation, you must rearrange the equation to isolate the desired variable (x). This can be done by using algebraic operations such as addition, subtraction, multiplication, and division.

What are some real-life applications of the Circular Current Equation?

The Circular Current Equation is used in various fields such as engineering, physics, and medicine. It is essential for designing and understanding the behavior of electromagnets, motors, and generators. It is also used in medical imaging techniques such as magnetic resonance imaging (MRI).

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