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Chapter 7: Problem 27

Ring in a solenoid ** An infinite solenoid with radius \(b\) has \(n\) turns per unit length. The current varies in time according to \(I(t)=I_{0} \cos \omega t\) (with positive defined as shown in Fig. 7.36). A ring with radius \(r

Short Answer

Expert verified
The induced current in the ring can be calculated using Faraday's law and the definition of magnetic flux. The time at which the force on a part of the ring is maximum can be found by differentiating the force with respect to time and setting the derivative equal to zero. The overall effect of the force on the ring can be discussed by analyzing the force on each segment of the ring.

Step by step solution

01

Calculate the Induced Current

Using Faraday's law of electromagnetic induction, the induced emf in the loop can be given by \( \varepsilon = - \frac{d \Phi }{dt} \), where \( \Phi \) is the magnetic flux through the ring. The flux \( \Phi = B \cdot A \), with B being the magnetic field and A being the area of the loop. Substitute the expressions for B and A into the equation and differentiate with respect to time to find \( \varepsilon \). Finally, using Ohm's law, the current I in the ring can be found by \( I=\frac{\varepsilon }{R} \).
02

Determine the Time for Maximum Force

The magnetic force on a small segment of the ring can be given by \( F= B \cdot I \cdot L \), where L is the length of the segment. Differentiate this force with respect to time and set the derivative equal to zero to find the time at which this force is maximum.
03

Discuss the Effect of the Force on the Ring

The force on each segment of the ring should be evaluated considering its direction (given by the right-hand rule for the cross product) in order to determine the overall effect of the force on the ring.

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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 of Electromagnetic Induction is fundamental in understanding how electric currents are induced in a conductor by changing magnetic fields. It's expressed by the formula:
  • \( \varepsilon = -\frac{d \Phi }{dt} \)
Here, \( \varepsilon \) is the electromotive force (emf) induced, and \( \Phi \) represents the magnetic flux.
The negative sign indicates Lenz's Law, which states that the direction of induced emf opposes the change in magnetic flux. Faraday's Law helps us find the induced current in the ring by taking into account the rate at which the magnetic field through the ring changes over time.
The change in magnetic field could be due to the current in the solenoid, \( I(t)=I_{0} \cos \omega t \), which fluctuates as time, \( t \), changes.
By differentiating \( \Phi \) with respect to time, we understand how the magnetic field changes and directly influence the induced current by combining with Ohm's Law.
Magnetic Flux
Magnetic Flux is a measure of the quantity of magnetism, considering the strength and extent of a magnetic field. It is given by the formula:
  • \( \Phi = B \cdot A \)
In this equation:
  • \( B \) is the magnetic field.
  • \( A \) is the area through which the magnetic field lines pass.
The magnetic flux through a given area provides the potential for inducing electric current via electromagnetic induction.
In the exercise, the magnetic flux is linked to the current \( I(t) \) flowing through the solenoid.
The flux determines how much magnetic field is 'felt' by the loop, thereby influencing the induced emf and current.
Understanding magnetic flux introduces a picture of how magnetic field lines intersect the loop, modifying the condition based on the solenoid's current pattern \( I(t) \) over time. This leads to changes in flux and subsequently, the induced current.
Ohm's Law
Ohm's Law is essential in linking the induced emf with the final current output in a circuit. It's summarized by the equation:
  • \( I = \frac{\varepsilon}{R} \)
Where:
  • \( I \) is the current in the circuit.
  • \( \varepsilon \) is the induced emf (from Faraday's law).
  • \( R \) represents the resistance of the circuit.
Ohm鈥檚 Law helps evaluate how efficiently the system (in this case, the ring) can convert induced voltage into electric current.
When analyzing the ring within the solenoid, using Ohm鈥檚 Law lets us calculate the induced current by dividing the emf value (found using Faraday's Law) with the resistance \( R \) of the ring.
This step helps in determining whether the current can drive the forces that influence physical effects such as the ring's movement, stretch, or spin due to its interaction with the solenoid's fluctuating magnetic field.

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Most popular questions from this chapter

Doubling a solenoid * (a) Two identical solenoids are connected end-to-end to make a solenoid of twice the length. By what factor is the selfinductance increased? The answer quickly follows from the formula for a solenoid's \(L\), but you should also explain in words why the factor is what it is. (b) Same question, but now with the two solenoids placed right on top of one another. (Imagine that one solenoid is slightly wider and surrounds the other.) They are connected so that the current flows in the same direction in each.

Current in a bottle \(*\) An ocean current flows at a speed of 2 knots (approximately \(1 \mathrm{~m} / \mathrm{s}\) ) in a region where the vertical component of the earth's magnetic field is \(0.35\) gauss. The conductivity of seawater in that region is 4 (ohm-m) \(^{-1}\). On the assumption that there is no other horizontal component of \(\mathbf{E}\) than the motional term \(\mathbf{v} \times \mathbf{B}\), find the density \(J\) of the horizontal electric current. If you were to carry a bottle of seawater through the earth's field at this speed, would such a current be flowing in it?

Two expressions for the energy (general) *** The task of Problem \(2.24\) was to demonstrate that two different expressions for the electrostatic energy, \(\int\left(\epsilon_{0} E^{2} / 2\right) d v\) and \(\int(\rho \phi / 2) d v\), are equivalent (as they must be, if they are both valid). The latter expression can quickly be converted to \(C \phi^{2} / 2\) in the case of oppositely charged conductors in a capacitor (see Exercise \(3.65\) ). The task of this problem is to demonstrate the analogous relation for the magnetic energy, that is, to show that if a circuit (of finite extent) with self-inductance \(L\) contains current \(I\), then \(\int\left(B^{2} / 2 \mu_{0}\right) d v\) equals \(L I^{2} / 2 .\) This is a bit trickier than the electrostatic case, so here are some hints: (1) a useful vector identity is \(\nabla \cdot(\mathbf{A} \times \mathbf{B})=\mathbf{B} \cdot(\nabla \times \mathbf{A})-\mathbf{A} \cdot(\nabla \times \mathbf{B}),(2)\) the vector potential and magnetic field satisfy \(\nabla \times \mathbf{A}=\mathbf{B}\), (3) \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}\), (4) \(\Phi=\int \mathbf{A} \cdot d l\) from Eq. (7.52), and (5) \(L\) is defined by \(\Phi=L I\).

Mutual-inductance symmetry \(* *\) In Section \(7.7\) we made use of the vector potential to prove that \(M_{12}=M_{21} .\) We can give a second proof, this time in the spirit of Exercise 3.64. Imagine increasing the currents in two circuits gradually from zero to the final values of \(I_{1 \mathrm{f}}\) and \(I_{2 \mathrm{f}}(\) "'f" for "final"). Due to the induced emfs, some external agency has to supply power to increase (or maintain) the currents. The final currents can be brought about in many different ways. Two possible ways are of particular interest. (a) Keep \(I_{2}\) at zero while raising \(I_{1}\) gradually from zero to \(I_{1 \mathrm{f}}\). Then raise \(I_{2}\) from zero to \(I_{2 \mathrm{f}}\) while holding \(I_{1}\) constant at \(I_{1 \mathrm{f}}\). (b) Carry out a similar program with the roles of 1 and 2 exchanged, that is, raise \(I_{2}\) from zero to \(I_{2} \mathrm{f}\) first, and so on.

L for a solenoid \(*\) Find the self-inductance of a long solenoid with radius \(r\), length \(\ell\), and \(N\) turns.
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