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Chapter 7: Problem 15

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.

Short Answer

Expert verified
Inside: \( E = -\frac{\mu_0 n s}{2} \frac{dI}{dt} \), Outside: \( E = -\frac{\mu_0 n a^2}{2s} \frac{dI}{dt} \) (both opposed to \( \hat{\phi} \)).

Step by step solution

01

Understanding the Problem

We need to find the electric field both inside and outside a long solenoid carrying a time-dependent current. The current's change induces an electric field due to Faraday's Law.
02

Apply Faraday's Law for Inside the Solenoid

Inside the solenoid (0 \leq s \leq a), we use Faraday's Law of Induction in integral form: \( \oint \mathbf{E} \cdot d\mathbf{l} = - \frac{d\Phi}{dt} \), where \( \Phi \) is the magnetic flux. The path of integration is a circle of radius \( s \).
03

Calculate Magnetic Flux Change Inside

The magnetic field inside the solenoid is uniform and given by \( B(t) = \mu_0 n I(t) \). The flux \( \Phi = B \times \text{area} = \mu_0 n I(t) \pi s^2 \). Differentiating with respect to time: \( \frac{d\Phi}{dt} = \mu_0 n \pi s^2 \frac{dI}{dt} \).
04

Determine Electric Field Inside

For a path length \( l = 2\pi s \), apply Faraday's Law: \( E \cdot 2\pi s = -\mu_0 n \pi s^2 \frac{dI}{dt} \). Solve for \( E \): \( E = -\frac{\mu_0 n s}{2} \frac{dI}{dt} \) with direction opposing \( \hat{\phi} \).
05

Apply Faraday's Law for Outside the Solenoid

Outside the solenoid (a < s), the magnetic flux inside the loop of radius \( s \) is \( \Phi = B \pi a^2 \) since the field is zero outside.
06

Calculate Electric Field Outside

Substitute \( \Phi = \mu_0 n I(t) \pi a^2 \) in Faraday's Law: \( E \cdot 2\pi s = -\mu_0 n \pi a^2 \frac{dI}{dt} \). Solve for \( E \): \( E = -\frac{\mu_0 n a^2}{2s} \frac{dI}{dt} \) with direction opposing \( \hat{\phi} \).

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Key Concepts

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

Faraday's Law
Faraday's Law is a fundamental principle of electromagnetism that describes how a changing magnetic field can induce an electric field. This concept is beautifully encapsulated in Faraday's Law of Induction, which in integral form is expressed as:
  • \( \oint \mathbf{E} \cdot d\mathbf{l} = - \frac{d\Phi}{dt} \)
This equation tells us that the line integral of the electric field \(\mathbf{E}\) around a closed path is equal to the negative rate of change of the magnetic flux \(\Phi\) through the surface enclosed by the path. Breaking it down:
  • The "line integral" of the electric field calculates the total electric effect along a path.
  • The "magnetic flux" \(\Phi\) is the product of the magnetic field \(B\) and the area it penetrates, aligning perpendicularly.
  • A negative sign indicates the induced electric field opposes the change in magnetic flux, consistent with Lenz's Law.
In our exercise, as the current \(I(t)\) within the solenoid changes over time, it alters the magnetic field inside the solenoid, thus affecting the magnetic flux. This variation induces an electric field both inside and outside the solenoid, which is precisely what Faraday's Law predicts.
Solenoid
A solenoid is an essential electrical component that consists of a tightly coiled wire. It serves as a simple way to create a strong, uniform magnetic field within its confines when an electric current passes through it. Here are some key points about solenoids:
  • The magnetic field inside a solenoid is almost uniform and parallel to the axis of the solenoid.
  • This field can be calculated using the formula \( B = \mu_0 n I \), where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current passing through the solenoid.
  • Outside the solenoid, the magnetic field is weak and considered negligible for ideal long solenoids.
In the exercise, the solenoid carries a time-dependent current \( I(t) \), indicating that the current must vary over time. This variation means that the magnetic field inside the solenoid will change, leading to an induced electric field according to Faraday's Law. Understanding the behavior of fields in and around solenoids is crucial for predicting and calculating electromagnetic effects.
Electric Field
An electric field is a region around a charged particle or changing magnetic field where another charged object experiences a force. Electric fields are vector fields, meaning they have both magnitude and direction, which affects the trajectory of charges within them. Here's more:
When induced by a changing magnetic field, as described by Faraday's Law, the electric field has specific characteristics:
  • Inside the solenoid, the electric field magnitude is determined by the equation \( E = -\frac{\mu_0 n s}{2} \frac{dI}{dt} \). The negative sign indicates the direction is such that it opposes the change in current, consistent with Lenz's law.

  • Outside the solenoid, the field is proportional to \( \frac{1}{s} \) and given by \( E = -\frac{\mu_0 n a^2}{2s} \frac{dI}{dt} \), again opposing the current's change.
  • The direction of the induced electric fields, both inside and outside, is in the azimuthal direction (around the solenoid), following the right-hand rule relative to the solenoid鈥檚 magnetic field.
Electric fields induced in this way are non-conservative; they depend on the path taken, a characteristic feature of time-varying magnetic fields. This means that the energy conversion processes in electromagnetic induction follow different rules than those for static fields, emphasizing the dynamic nature of electromagnetism.

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Most popular questions from this chapter

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.

The magnetic field of an infinite straight wire cartying a steady current \(I\) can be obtained from the displacement current term in the Amp猫re/Maxwell law, as follows: Picture the current as consisting of a uniform line charge \(\lambda\) moving along the \(z\) axis at speed \(v\) (so that \(I=\lambda v\) ), with a tiny gap of length \(\epsilon\), which reaches the origin at time \(t=0 .\) In the next instant (up to \(t=\epsilon / v\) ) there is no real current passing through a circular Amperian loop in the \(x y\) plane, but there is a displacement current, due to the "missing" charge in the gap. (a) Use Coulomb's law to calculate the \(z\) component of the electric field, for points in the \(x y\) plane a distance \(s\) from the origin, due to a scgment of wire with uniform density \(-\lambda\) exiending from \(z_{1}=v t-\epsilon\) to \(z_{2}=v t .\) (b) Determine the flux of this clectric field through a circle of radius \(a\) in the \(x y\) plane. (c) Find the displacement current through this circle. Show that \(I_{d}\) is equal to \(I\), in the limit as the gap width \((\epsilon)\) goes to zero. [For a slightly different approach to the same problem, see W. K. Terry, Am. J. Phys. 50. 742 (1982).]

'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.]

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.]

A familiar demonstration of superconductivity (Prob. \(7.42\) ) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. \({ }^{19}\) Treat the magnet as a perfect dipole \(\mathrm{m}\), a hcight \(z\) above the origin (and constrained to point in the \(z\) direction), and pretend that the superconductor occupies the entire half-space below the \(x y\) plane. Because of the Meissner cffect, \(\mathbf{B}=0\) for \(z \leq 0\), and since \(\mathbf{B}\) is divergenceless, the normal \((z)\) component is continuous, so \(B_{z}=0\) just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at \(-z\), as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region \(z>0 .\) (a) Which way should the image dipole point \((+z\) or \(-z)\) ? (b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to \(M g\) (where \(M\) is the mass of the magnet) to determine the height \(h\) at which the magnet will "float."
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