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A square loop of wire (side \(a\) ) lies on a table, a distance \(s\) from a very long straight wire, which carries a current \(I\), as shown in Fig. \(7.17\). (a) Find the flux of \(\mathrm{B}\) through the loop. (b) If someone now pulls the loop directly away from the wire, at speed \(v\), what emf is generated? In what direction (clockwise or counterclockwise) does the current flow? (c) What if the loop is puiled to the right at speed \(v\), instead of away?

Short Answer

Expert verified
(a) \( \Phi = \frac{\mu_0 I a}{2\pi} \ln \left(\frac{s+a}{s}\right) \) (b) EMF is \( \frac{\mu_0 I a v}{2\pi (s+vt)} \), counterclockwise. (c) EMF is zero.

Step by step solution

01

Understand the Magnetic Field due to the Wire

The magnetic field created by a long, straight wire carrying a current \( I \) is given by Ampère's Law. At a distance \( r \) from the wire, the magnetic field strength \( B \) is \( B = \frac{\mu_0 I}{2 \pi r} \), where \( \mu_0 \) is the permeability of free space.
02

Calculate the Flux through the Square Loop

The magnetic flux \( \Phi \) through an area is the integral of the magnetic field over that area. For this square loop: \[\Phi = \int_{s}^{s+a} \int_{0}^{a} B \, dx \, dr = \int_{s}^{s+a} \left( \frac{\mu_0 I}{2\pi r} \right) a \, dr. \] Calculating this integral gives us: \[\Phi = \frac{\mu_0 I a}{2\pi} \ln \left(\frac{s+a}{s}\right).\]
03

Apply Faraday's Law to Find EMF when Pulled Away

Faraday's Law states that the induced electromotive force (EMF) in any closed circuit is equal to the rate of change of magnetic flux through the circuit. If the loop moves away at speed \(v\): \[ \text{EMF} = -\frac{d\Phi}{dt} = -\frac{\mu_0 I a}{2\pi} \left(-\frac{v}{s+vt}\right).\] Simplifying, we have \( \text{EMF} = \frac{\mu_0 I a v}{2\pi (s+vt)} \). The negative sign indicates the direction of EMF opposing the change, resulting in counterclockwise current.
04

EMF When Loop is Pulled Parallel to the Wire

When the loop is moved parallel at speed \(v\), the distance from the wire does not change, hence the flux through the loop remains constant over time. This means \( \frac{d\Phi}{dt} = 0 \), and therefore, the EMF generated is zero.

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Key Concepts

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

Magnetic Field
A magnetic field is an invisible force field that surrounds magnetic materials and electric currents. It is a vector field, meaning it has both magnitude and direction. The source of a magnetic field can be examined using Ampère's Law, which helps calculate the strength of the magnetic field due to a current flowing through a wire. The formula given is \( B = \frac{\mu_0 I}{2 \pi r} \), where \( B \) is the magnetic field strength at a distance \( r \) from a long, straight wire carrying a current \( I \).

The presence of a magnetic field explains why two magnets or current-carrying wires experience force when placed in proximity to each other. The field lines that represent the magnetic field circulate around the wire in concentric circles. The direction can be determined using the right-hand rule: if you point your thumb in the direction of the current, the direction in which your fingers curl represents the direction of the magnetic field loops.
Faraday's Law
Faraday's Law of electromagnetic induction is a fundamental principle that explains how a change in magnetic flux through a loop of wire induces an electromotive force (EMF) in the loop. It is expressed mathematically as \( \text{EMF} = -\frac{d\Phi}{dt} \), where \( \Phi \) is the magnetic flux.

Magnetic flux is the product of the magnetic field, the area it penetrates, and the cosine of the angle between the magnetic field and the perpendicular to the surface. When the loop is pulled away from the wire in our scenario, the magnetic flux changes because the changing position alters the amount of the field passing through the loop. This alteration in flux induces an EMF, which is calculated by the rate of change of this flux.

The negative sign in Faraday's Law signifies Lenz's Law, which states that the induced EMF will always work to oppose the change in flux. If you pull the loop away, the current induced flows counterclockwise, opposing the reduction in magnetic field lines through the loop.
Ampère's Law
Ampère’s Law is a fundamental tool for calculating the magnetic field created by an electric current. It is particularly useful for situations with high symmetry and involves integrating the magnetic field around a closed loop path. In simple terms, it relates the integrated magnetic field around a closed loop to the electric current passing through the loop.

In our example, a long, straight wire carrying current \( I \) produces a magnetic field \( B \) that decreases with distance \( r \) from the wire. Using Ampère's Law, we derived that \( B = \frac{\mu_0 I}{2 \pi r} \). This equation shows that the strength of the magnetic field decreases as you move further from the wire. The direction of the magnetic field around the wire can be visualized as concentric circles, and its strength can be calculated in regions where the setup is symmetrical.
Magnetic Flux
Magnetic flux quantifies the total magnetic field that passes through a given area. It is the dot product of magnetic field \( B \) and the area vector \( A \), represented as \( \Phi = B \cdot A \). In more detail, it is calculated by integrating the magnetic field over an area:

\[ \Phi = \int B \cdot dA \]

For our exercise, when calculating the magnetic flux through the square loop, it involves integrating the magnetic field over the entire loop area, with factors such as distance from the wire and loop dimensions significantly affecting the result. The flux specifically is calculated using the derived expression: \( \Phi = \frac{\mu_0 I a}{2 \pi} \ln \left(\frac{s+a}{s}\right) \).

This describes how the proximity to the wire and the area of the loop interact with the magnetic field created by the current. Whenever the loop moves, either further away or parallel, the amount of flux through the loop may change, influencing the induced EMF, as explained by Faraday's Law.

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Most popular questions from this chapter

A long solenoid with radius \(a\) and \(n\) turns per unit length carries a time- dependent current \(I(t)\) in the \(\hat{\phi}\) direction. Find the electric field (magnitude and direction) at a distance \(s\) from the axis (both inside and outside the solenoid), in the quasistatic approximation.

Two coils are wrapped around a cylindrical form in such a way that the same flux passes through every turn of both coils. (In practice this is achieved by inserting an iron core through the cylinder; this has the effect of concentrating the flux.) The "primary" coil has \(N_{1}\) turns and the secondary has \(N_{2}\) (Fig. 7.54). If the current \(I\) in the primary is changing. show that the emf in the secondary is given by $$\frac{\mathcal{E}_{2}}{\mathcal{E}_{1}}=\frac{N_{2}}{N_{1}}$$ where \(\mathcal{E}_{1}\) is the (back) emf of the primary. [This is a primitive transformer-a device for raising or lowering the emf of an alternating current source. By choosing the appropriate number of turns, any desired secondary emf can be obtained. If you think this violates the conservation of energy, check out Prob. 7.54.]

'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.]

Find the self-inductance per unit length of a long solenoid, of radius \(R\), carrying \(n\) turns per unit length.

As a lecture demonstration a short cylindrical bar magnet is dropped down a vertical aluminum pipe of slightly larger diameter, about 2 meters long. It takes several seconds to emerge at the bottom, whereas an otherwise identical piece of unmagnetized iron makes the trip in a fraction of a second. Explain why the magnet falls more slowly.

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