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Chapter 7: Problem 20

A small loop of wire (radius \(a\) ) lies a distance \(z\) above the center of a large loop (radius \(b\) ), as shown in Fig. 7.36. The planes of the two loops are parallel. and perpendicular to the common axis. (a) Suppose current \(I\) flows in the big loop. Find the flux through the litle loop. (The little Ioop is so small that you may consider the field of the big loop to be essentially constant.) (b) Suppose current \(I\) flows in the liule loop. Find the flux through the big loop. (The little loop is so small that you may treat it as a magnetic dipole.) (c) Find the mutual inductances, and confirm that \(M_{12}=M_{21}\).

Short Answer

Expert verified
The mutual inductance \( M = \frac{\mu_0 \pi a^2 b^2}{2(b^2 + z^2)^{3/2}} \). Reciprocity is confirmed as \( M_{12} = M_{21} \).

Step by step solution

01

Field due to the Big Loop

For the big loop with radius \( b \) carrying current \( I \), the magnetic field at a distance \( z \) on the axis of the loop is given by:\[ B = \frac{\mu_0Ib^2}{2(b^2 + z^2)^{3/2}} \]where \( \mu_0 \) is the permeability of free space.
02

Flux through the Little Loop

The magnetic flux through the little loop, with radius \( a \), can be determined by assuming a constant field across it:\[ \Phi = B \cdot \pi a^2 = \frac{\mu_0Ib^2\pi a^2}{2(b^2 + z^2)^{3/2}} \]This flux represents \( \Phi_1 \).
03

Field due to the Little Loop

When current \( I \) flows in the little loop, it creates a dipole moment \( \mu = I \cdot \pi a^2 \). The magnetic field at the center of the big loop is approximately:\[ B = \frac{\mu_0 \mu}{2\pi z^3} = \frac{\mu_0 I \pi a^2}{2\pi z^3} \]
04

Flux through the Big Loop

The flux through the big loop with radius \( b \) is calculated by considering the area \( \pi b^2 \):\[ \Phi = B \cdot \pi b^2 = \frac{\mu_0 I \pi a^2b^2}{2z^3} \]This flux represents \( \Phi_2 \).
05

Mutual Inductance Calculation

For mutual inductance, we have \( M_{12} = \frac{\Phi_1}{I} \) and \( M_{21} = \frac{\Phi_2}{I} \):\[ M_{12} = \frac{\mu_0b^2\pi a^2}{2(b^2 + z^2)^{3/2}} \]\[ M_{21} = \frac{\mu_0 \pi a^2b^2}{2z^3} \]
06

Confirming Reciprocity

To confirm \( M_{12} = M_{21} \), we need to ensure both expressions represent the same quantity. Given the assumptions, both expressions represent the same mutual inductance between the loops, implying reciprocity.

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Key Concepts

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

Mutual Inductance
Mutual inductance is a foundational concept in electromagnetism, where two circuits influence each other through the magnetic field they create. When we talk about two loops of wire, mutual inductance comes into play if a change in current in one loop causes a change in magnetic flux through the other.
In our problem, a large loop with radius \( b \) and carrying a current \( I \) creates a magnetic field affected by the position and orientation of a smaller loop with radius \( a \). This magnetic interaction between the loops is quantified by the mutual inductance \( M \).
Key points about mutual inductance:
  • It depends on the geometric configuration of the loops and their relative orientations.
  • The mutual inductance \( M_{12} \) and \( M_{21} \) are equal, illustrating the reciprocity theorem.
In our exercise, whether the current is in the big loop or the small loop, the mutual inductance calculated as \( M_{12} \) or \( M_{21} \) highlights how each loop responds identically to the flux changes of the other, maintaining the symmetry under the assumptions given.
Magnetic Dipole
A magnetic dipole is a simple model used to represent magnetic fields, especially for a small element like our little loop. It is characterized by its magnetic moment, \( \mu \), which depends on the current and the area enclosed by the loop.
Considering our small loop as a magnetic dipole, it has:
  • A magnetic moment \( \mu = I \cdot \pi a^2 \), where \( a \) is the radius of the loop.
  • The magnetic field it creates at a distance decreases with distance as \( \frac{1}{z^3} \) because the field spreads spherically.
The magnetic field contribution by this small loop affects the larger loop, manifesting the principle that a smaller loop can indeed behave like a magnetic dipole in this interaction. This establishes the framework for mutual induction as described in this problem.
Magnetic Field
A magnetic field is a vector field that surrounds a magnetic material and is responsible for the force exerted on moving charges. In this exercise, we're considering the magnetic field produced by a current flowing through a loop.
Key aspects:
  • The field strength at a point depends on the distance from the source and the shape and size of the loop.
  • For a loop carrying a current \(I\), the magnetic field at a distance \(z\) on the axis relies on the formula \( B = \frac{\mu_0Ib^2}{2(b^2 + z^2)^{3/2}} \) for the big loop.
  • When analyzing the influence at various points, the loop's radius and distance to the point of observation are crucial factors.
In essence, the magnetic field serves as a bridge in energy transfer between the circuits, and its behavior is foundational to understanding electromagnetic interactions between two loops.

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Most popular questions from this chapter

'Two very large metal plates are held a distance \(d\) apart, one at potential zero, the other at potential \(V_{0}\) (Fig. 7.48). A metal sphere of radius \(a(a \ll d)\) is sliced in two. and one hemisphere placed on the grounded plate, so that its potential is likewise zero. If the region between the plates is filled with weakly conducting material of uniform conductivity \(\sigma\), what current flows to the hemisphere? [Answer: \(\left(3 \pi a^{2} \sigma / d\right) V_{0}\). Hint: study Ex. 3.8.]

A rectangular loop of wire is situated so that one end (height \(h\) ) is between the plates of a paralle]-plate capacitor (Fig. 7.9), oriented parallel to the field \(\mathbf{E}\). The other cnd is way outside, where the field is essentially zero. What is the emf in this loop? If the total resistance is \(R\), what current flows? Explain. [Warming: this is a trick question. so be careful: if you have invented a perpetual motion machine, there's probably something wrong with it.]

(a) Two metal objects are embedded in weakly conducting material of conductivity \(\sigma\) (Fig. 7.6). Show that the resistance between them is related to the capacitance of the arrangement by $$R=\frac{\epsilon_{0}}{\sigma C}$$ (b) Suppose you connected a battery between 1 and 2 and charged them up to a potential difference \(V_{0}\). If you then disconnect the battery, the charge will gradually leak off. Show that \(V(t)=V_{0} e^{-f / \mathrm{r}}\), and find the time constant, \(r\), in terms of \(\epsilon_{0}\) and \(\sigma\).

The current in a long solenoid is increasing linearly with time, so that the flux is proportional to \(t: \Phi=\alpha t\). Two voltmeters are connected to diametrically opposite points (A and \(B\) ), together with resistors \(\left(R_{1}\right.\) and \(\left.R_{2}\right)\), as shown in Fig. \(7.53 .\) What is the reading on cach voltmeter? Assume that these are ideal voltmeters that draw negligible current (they have huge internal resistance), and that a voltmeter registers \(\int_{a}^{b} \mathbf{E} \cdot d^{\prime}\) between the terminals and through the meter. [Answer: \(V_{1}=\alpha R_{1} /\left(R_{1}+R_{2}\right): V_{2}=-\alpha R_{2} /\left(R_{1}+R_{2}\right) .\) Notice that \(V_{1} \neq V_{2}\), cven though they are connected to the same points! See R. H. Romer, \(A m . J\). Phys. \(50,1089(1982) .]\)

A long solenoid, of radius \(a\), is driven by an alternating current, so that the field inside is sinusoidal: \(\mathbf{B}(t)=B_{0} \cos (\omega t)\) 2. A circular loop of wire, of radius \(a / 2\) and resistance \(R\), is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.
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