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Ttwo parallel loops of wire having a common axis. The smaller loop (radius \(r\) ) is above the larger loop (radius \(R\) ) by a distance \(x \geqslant R\). Consequently, the magnetic field due to the counterclockwise current \(i\) in the larger loop is nearly uniform throughout the smaller loop. Suppose that \(x\) is increasing at the constant rate \(d x / d t=v .\) (a) Find an expression for the magnetic flux through the area of the smaller loop as a function of \(x\).

Short Answer

Expert verified
The magnetic flux through the smaller loop, as a function of \( x \), is \( \Phi(x) = \frac{\mu_0 i R^2 \pi r^2}{2(x^2 + R^2)^{3/2}} \).

Step by step solution

01

Understanding Magnetic Flux

Magnetic flux \( \Phi \) is defined as the product of the magnetic field \( B \) and the area \( A \) through which the field lines pass, i.e., \( \Phi = B \cdot A \). For a uniform magnetic field and a flat loop, this simplifies the calculation.
02

Magnetic Field of the Larger Loop

The magnetic field \( B \) at distance \( x \) from a loop with radius \( R \) carrying current \( i \) is given by the formula \( B(x) = \frac{\mu_0 i R^2}{2(x^2 + R^2)^{3/2}} \), where \( \mu_0 \) is the permeability of free space.
03

Calculating the Area of the Smaller Loop

The area \( A \) of the smaller loop can be expressed as \( A = \pi r^2 \), where \( r \) is the radius of the smaller loop.
04

Expression for Magnetic Flux Through the Smaller Loop

Using the magnetic field expression from Step 2 and the area from Step 3, the magnetic flux \( \Phi \) through the smaller loop is \( \Phi = B(x) \cdot A = \frac{\mu_0 i R^2 \pi r^2}{2(x^2 + R^2)^{3/2}} \).

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Key Concepts

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

Magnetic Field
When we think of a magnetic field, imagine the invisible lines of force around a magnet. The magnetic field is generated by electric currents, such as the flow of electricity through a wire. It is a vector field, meaning it has both a direction and a magnitude. In this exercise, the magnetic field is created by a current flowing through a circular loop of wire. The further you are from the center of this loop, the weaker the magnetic field becomes. That is why the distance from the loop, denoted as \( x \), plays a crucial role.

The magnetic field created by a loop at a point on its axis can be calculated using the formula:\[B(x) = \frac{\mu_0 i R^2}{2(x^2 + R^2)^{3/2}}\]Here:
  • \( B(x) \) is the magnetic field at distance \( x \).
  • \( R \) is the radius of the loop.
  • \( i \) is the current passing through the loop.
  • \( \mu_0 \) is the permeability of free space, a constant that describes how a magnetic field can traverse through space.
Understanding the concept of the magnetic field and how it diminishes with distance is vital for solving problems related to loops of wire and their fields.
Loops of Wire
A loop of wire isn't just a random circle of metal; it's a fundamental component in electromagnetism. When electric current flows through the wire, it creates a magnetic field around the loop. In this problem, two loops of wire are described—one larger and one smaller, with the smaller loop positioned above the larger one.

The smaller loop's magnetic field interaction with the larger one is significant. If the magnetic field across the smaller loop is nearly uniform, the setup becomes easier to analyze. The magnetic flux through the smaller loop is essentially the product of this uniform magnetic field and the loop's area.
  • Radius of the smaller loop, \( r \), determines its area.
  • The area \( A \) of this loop is \( \pi r^2 \).
  • The positioning and size of the loops affect the overall magnetic field experienced by the smaller loop.
As you study loops of wire, recognize how their dimensions and relative positions influence the resulting magnetic interactions.
Electromagnetism
Electromagnetism is the branch of physics that studies the relationship between electric currents and magnetic fields. It’s fundamentally about how electricity and magnetism interact. In this exercise, the relationship is highlighted by calculating the magnetic flux \( \Phi \) through the smaller wire loop as a function of its distance \( x \) from the larger loop.

Magnetic flux is the measure of the total magnetic field passing through a certain area, and is expressed as:\[\Phi = B \cdot A\]In our scenario:
  • \( B \) is the magnetic field created by the larger loop.
  • \( A \) is the area of the smaller loop.
The expression for the flux in this problem simplifies to:\[\Phi = \frac{\mu_0 i R^2 \pi r^2}{2(x^2 + R^2)^{3/2}}\]This equation reflects how the magnetic field generated by the larger loop's current influences the smaller loop, and helps in understanding the broader landscape of electromagnetism—where circuits, fields, and forces coalesce to create a magnetic interaction you can mathematically analyze.

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Most popular questions from this chapter

Figure \(30-56\) shows a copper strip of width \(W=16.0 \mathrm{~cm}\) that has been bent to form a shape that consists of a tube of radius \(R=1.8 \mathrm{~cm}\) plus two parallel flat extensions. Current \(i=35 \mathrm{~mA}\) is distributed uniformly across the width so that the tube is effectively a one-turn solenoid. Assume that the magnetic field outside the tube is negligible and the field inside the tube is uniform. What are (a) the magnetic field magnitude inside the tube and (b) the inductance of the tube (excluding the flat extensions)?

A solenoid having an inductance of \(6.30 \mu \mathrm{H}\) is connected in series with a \(1.20 \mathrm{k} \Omega\) resistor. (a) If a \(14.0 \mathrm{~V}\) battery is connected across the pair, how long will it take for the current through the resistor to reach \(80.0 \%\) of its final value? (b) What is the current through the resistor at time \(t=1.0 \tau_{L} ?\)

A metal rod is forced to move with constant velocity \(\vec{v}\) along two parallel metal rails, connected with a strip of metal at one end. A magnetic field of magnitude \(B=0.350 \mathrm{~T}\) points out of the page. (a) If the rails are separated by \(L=25.0 \mathrm{~cm}\) and the speed of the rod is \(55.0 \mathrm{~cm} / \mathrm{s}\), what emf is generated? (b) If the rod has a resistance of \(18.0 \Omega\) and the rails and connector have negligible resistance, what is the current in the rod? (c) At what rate is energy being transferred to thermal energy?

Shows a rod of length \(L=10.0 \mathrm{~cm}\) that is forced to move at constant speed \(v=5.00\) \(\mathrm{m} / \mathrm{s}\) along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance \(0.400 \Omega\); the rest of the loop has negligible resistance. A current \(i=100 \mathrm{~A}\) through the long straight wire at distance \(a=10.0 \mathrm{~mm}\) from the loop sets up a (nonuniform) magnetic field through the loop. Find the (a) emf and (b) current induced in the loop. (c) At what rate is thermal energy generated in the rod? (d) What is the magnitude of the force that must be applied to the rod to make it move at constant speed? (e) At what rate does this force do work on the rod?

A uniform magnetic field \(\vec{B}\) is perpendicular to the plane of a circular loop of diameter \(10 \mathrm{~cm}\) formed from wire of diameter \(2.5 \mathrm{~mm}\) and resistivity \(1.69 \times 10^{-8} \Omega \cdot \mathrm{m}\). At what rate must the magnitude of \(\vec{B}\) change to induce a \(10 \mathrm{~A}\) current in the loop?

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