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Chapter 29: Problem 59

A 100 -turn solenoid of length \(8 \mathrm{~cm}\) and radius \(6 \mathrm{~mm}\) carries a current of 0.4 A from right to left. The current is then reversed so that it flows from left to right. By how much does the energy stored in the magnetic field inside the solenoid change?

Short Answer

Expert verified
Answer: The change in energy stored in the magnetic field inside the solenoid after the current is reversed is zero.

Step by step solution

01

Calculate the cross-sectional area of the solenoid

We are given the radius of the solenoid (\(r = 6 \mathrm{mm}\)). To find the cross-sectional area \((A)\), we'll use the formula: \(A = \pi r^2\) Once we have the area, we will use it to determine the inductance of the solenoid.
02

Calculate the inductance of the solenoid

With the cross-sectional area calculated, we can now determine the inductance (\(L\)) of the solenoid using the formula: \(L = \dfrac{\mu_0 N^2 A}{l}\) We are given all other necessary values: \(N = 100\) \(l = 8 \mathrm{cm} = 0.08 \mathrm{m}\) \(\mu_0 = 4\pi \times 10^{-7} \mathrm{T\cdot m/A}\)
03

Calculate the energy stored in the magnetic field before and after the current reversal

We will now calculate the energy stored in the magnetic field (\(U\)) before and after the current reversal using the inductance (\(L\)) we calculated in step 2 and the given current values (\(I\)). Since the magnitude of the current is the same before and after the reversal, the energy stored in the magnetic field for both cases will be the same: \(U_1 = U_2 = \dfrac{1}{2}L I^2\) Where: \(U_1\) - Energy stored before the current reversal \(U_2\) - Energy stored after the current reversal
04

Calculate the change in energy stored in the magnetic field

Finally, we can find the change in energy stored in the magnetic field by finding the difference between the energy stored before and after the current reversal: \(\Delta U = |U_2 - U_1|\) Since both energies are equal, the change in energy stored in the magnetic field 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 field around a magnet or current-carrying wire. It is represented by magnetic field lines showing the direction and strength of the field.
  • The strength of the magnetic field is measured in Tesla (T).
  • The magnetic field direction is given by the right-hand rule: point your thumb in the direction of the current, and your fingers curl in the direction of the field lines.
Understanding how magnetic fields work helps us see how currents in solenoids create them.
Solenoid
A solenoid is a coil of wire that generates a magnetic field when an electric current passes through it. It acts like a magnet and is widely used in various applications.
  • When current flows through the solenoid, it creates a magnetic field inside it.
  • The strength of this field depends on factors like the number of turns in the coil and the current strength.
  • Solenoids are useful in creating controlled magnetic fields for devices like relays and electromagnets.
In the problem, turning the current direction changes the magnetic field orientation.
Inductance
Inductance is a property of a coil that measures its ability to store energy in a magnetic field. It depends on the coil's physical characteristics.
  • It is measured in Henry (H).
  • For a solenoid, inductance is calculated using the formula: \(L = \frac{\mu_0 N^2 A}{l}\), where \(\mu_0\) is the permeability of free space, \(N\) is the number of turns, \(A\) is the cross-sectional area, and \(l\) is the length.
  • Higher inductance means the solenoid can store more energy.
Understanding inductance helps in analyzing how much energy the solenoid can store.
Energy Storage
The energy in a magnetic field gets stored when a current flows through an inductor, like a solenoid. This energy can be calculated with the inductance and current value.
  • The energy stored \(U\) in an inductor is given by \(U = \frac{1}{2}L I^2\).
  • The energy depends on both the inductance \(L\) and the current \(I\).
  • Reversing the current does not change the energy, as the formula considers the magnitude of the current.
  • Thus, if the current is reversed, the stored energy remains unchanged, as explained in the original problem's solution.
This concept highlights the enduring relationship between current flow and energy storage in magnetic fields.

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Most popular questions from this chapter

A popular demonstration of eddy currents involves dropping a magnet down a long metal tube and a long glass or plastic tube. As the magnet falls through a tube, there is changing flux as the magnet falls toward or away from each part of the tube. a) Which tube has the larger voltage induced in it? b) Which tube has the larger eddy currents induced in it?

A rectangular wire loop (dimensions of \(h=15.0 \mathrm{~cm}\) and \(w=8.00 \mathrm{~cm}\) ) with resistance \(R=5.00 \Omega\) is mounted on a door. The Earth's magnetic field, \(B_{\mathrm{E}}=2.6 \cdot 10^{-5} \mathrm{~T}\), is uniform and perpendicular to the surface of the closed door (the surface is in the \(x z\) -plane). At time \(t=0,\) the door is opened (right edge moves toward the \(y\) -axis) at a constant rate, with an opening angle of \(\theta(t)=\omega t,\) where \(\omega=3.5 \mathrm{rad} / \mathrm{s}\) Calculate the direction and the magnitude of the current induced in the loop, \(i(t=0.200 \mathrm{~s})\).

When a magnet in an MRI is abruptly shut down, the magnet is said to be quenched. Quenching can occur in as little as \(20.0 \mathrm{~s}\). Suppose a magnet with an initial field of \(1.20 \mathrm{~T}\) is quenched in \(20.0 \mathrm{~s},\) and the final field is approximately zero. Under these conditions, what is the average induced potential difference around a conducting loop of radius \(1.00 \mathrm{~cm}\) (about the size of a wedding ring) oriented perpendicular to the field?

A steel cylinder with radius \(2.5 \mathrm{~cm}\) and length \(10.0 \mathrm{~cm}\) rolls without slipping down a ramp that is inclined at \(15^{\circ}\) above the horizontal and has a length (along the ramp) of \(3.0 \mathrm{~m} .\) What is the induced potential difference between the ends of the cylinder at the bottom of the ramp, if the surface of the ramp points along magnetic north?

What is the inductance in a series \(\mathrm{RL}\) circuit in which \(R=3.00 \mathrm{k} \Omega\) if the current increases to one half of its final value in \(20.0 \mu \mathrm{s} ?\)
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