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Chapter 28: Problem 35

Two concentric circular loops of wire lie on a tabletop, one inside the other. The inner wire has a diameter of \(20.0 \mathrm{~cm}\) and carries a clockwise current of \(12.0 \mathrm{~A}\), as viewed from above, and the outer wire has a diameter of \(30.0 \mathrm{~cm} .\) What must be the magnitude and direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

Short Answer

Expert verified
The current in the outer wire must be \(8.00 \mathrm{A}\) and flow counter-clockwise when viewed from above to counteract the magnetic field produced by the current in the inner wire.

Step by step solution

01

Calculate the magnetic field of the inner wire

First calculate the magnetic field due to the current in the inner wire. Using the formula \(B = \mu I / 2r\), where \(B\) is the magnetic field strength, \(\mu\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{T m/A}\)), \(I\) is the current (12.0 A), and \(r\) is the radius of the loop (half of the diameter of 0.20 m). This gives an initial magnetic field \(B_{\text{inner}} = (4\pi \times 10^{-7} \mathrm{T m/A} \times 12.0 \mathrm{A}) / (2 \times 0.10 \mathrm{m})\).
02

Solve for the current in the outer wire

The magnetic field due to the current in the outer wire must be equal and opposite to the magnetic field due to the inner wire so that the total field is zero. This condition can be written as \(-B_{\text{inner}} = \mu I_{\text{outer}} / 2r_{\text{outer}}\), where \(B_{\text{inner}}\) is the magnetic field due to the inner wire calculated in the previous step, \(I_{\text{outer}}\) is the current in the outer wire that we are trying to find, and \(r_{\text{outer}}\) is the radius of the outer wire (half of the diameter of 0.30 m). By rearranging this equation we can solve for the current in the outer wire: \(I_{\text{outer}} = -2 B_{\text{inner}} r_{\text{outer}} / \mu\). The negative sign indicates that the current must flow in the opposite direction to the current in the inner wire.
03

Substitute the values and compute the current

Substitute the values \(B_{\text{inner}}\), \(r_{\text{outer}}\), and \(\mu\) into the equation from step 2 to calculate the current in the outer wire: \(I_{\text{outer}} = -2 B_{\text{inner}} r_{\text{outer}} / \mu = -2 B_{\text{inner}} (0.15 \mathrm{m}) / 4\pi \times 10^{-7} \mathrm{T m/A}\). This will give the magnitude and direction of the current in the outer wire.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concentric Circular Loops
Concentric circular loops are simply two or more circles that share the same center point but have different radii. In the context of this problem, the loops are formed by wires that carry electric current. The key characteristic of these loops is that they do not touch each other, although they share a central axis.

For our exercise, this means that the inner loop has a radius of 10 cm, from a diameter of 20 cm, and the outer loop has a radius of 15 cm from a diameter of 30 cm. The magnetic fields produced by currents in these loops can interact with each other. Understanding how these fields interact is crucial in solving the problem.
  • The center of both loops is the same, facilitating their interaction.
  • The loops are positioned flat on a surface such as a tabletop, as viewed from above.
Knowing the physical setup of these loops helps visualize the magnetic interaction and is a critical part of approaching the exercise.
Magnetic Field Cancellation
When two magnetic fields are equal in magnitude but opposite in direction, they effectively cancel each other out at a given point. This principle of magnetic field cancellation is essential when dealing with multiple current-carrying loops or wires.

For our problem, the goal is to set up the current in the outer loop such that its magnetic field cancels out the magnetic field produced by the inner loop at the common center.
  • The inner loop's current generates a clockwise magnetic field.
  • To achieve zero net magnetic field at the center, the outer loop needs to generate a field of equal strength but in the opposite direction (counterclockwise).
This condition ensures that the fields from both loops combine to a total of zero, effectively neutralizing each other at the center.
Biot-Savart Law
The Biot-Savart Law provides a mathematical approach to determining the magnetic field created by a current-carrying segment of wire. It is particularly useful in calculating the field produced by circular wires or loops.

The law states that the magnetic field \(B\) at a point is proportional to the current \(I\), and inversely proportional to the distance from the wire, denoted as \(r\). The formula used in the solution is a simplified version: \(B = \frac{\mu I}{2r}\), where \(\mu\) represents the permeability of free space.
  • This law helps derive the inner loop's magnetic field by using its current and radius.
  • With the Biot-Savart's formula, we equate the outer loop's field and solve for the required current.
Applying this law is vital to solving the problem, understanding the manner in which magnetic fields behave around current loops.

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Most popular questions from this chapter

A long, straight, solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}}\). The current density, although symmetric about the cylinder axis, is not constant but varies according to the relationship $$ \begin{array}{rlr} \overrightarrow{\boldsymbol{J}} & =\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ & =\mathbf{0} \quad & \text { for } r \geq a \end{array} $$ where \(a\) is the radius of the cylinder, \(r\) is the radial distance from the cylinder axis, and \(I_{0}\) is a constant having units of amperes. (a) Show that \(I_{0}\) is the total current passing through the entire cross section of the wire. (b) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \geq a\). (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. (d) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \leq a\). How do your results in parts (b) and (d) compare for \(r=a ?\)

You use a teslameter (a Hall-effect device) to measure the magnitude of the magnetic field at various distances from a long, straight, thick cylindrical copper cable that is carrying a large constant current. To exclude the earth's magnetic field from the measurement, you first set the meter to zero. You then measure the magnetic field \(B\) at distances \(x\) from the surface of the cable and obtain these data: $$ \begin{array}{l|lllll} x(\mathrm{~cm}) & 2.0 & 4.0 & 6.0 & 8.0 & 10.0 \\ \hline B(\mathrm{mT}) & 0.406 & 0.250 & 0.181 & 0.141 & 0.116 \end{array} $$ (a) You think you remember from your physics course that the magnetic field of a wire is inversely proportional to the distance from the wire. Therefore, you expect that the quantity \(B x\) from your data will be constant. Calculate \(B x\) for each data point in the table. Is \(B x\) constant for this set of measurements? Explain. (b) Graph the data as \(x\) versus \(1 / B\). Explain why such a plot lies close to a straight line. (c) Use the graph in part (b) to calculate the current \(I\) in the cable and the radius \(R\) of the cable.

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be \(55.0 \mathrm{~cm}\) long and \(2.80 \mathrm{~cm}\) in diameter. What current will you need to produce the necessary field?

Long, straight conductors with square cross sections and each carrying current \(I\) are laid side by side to form an infinite current sheet (Fig. \(\mathbf{P 2 8 . 6 9}\) ). The conductors lie in the \(x y\) -plane, are parallel to the \(y\) -axis, and carry current in the \(+y\) -direction. There are \(n\) conductors per unit length measured along the \(x\) -axis. (a) What are the magnitude and direction of the magnetic field a distance \(a\) below the current sheet? (b) What are the magnitude and direction of the magnetic field a distance \(a\) above the current sheet?

A \(15.0-\mathrm{cm}\) -long solenoid with radius \(0.750 \mathrm{~cm}\) is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the center of the solenoid.
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