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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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Most popular questions from this chapter

Chapter 14 discussed damped harmonic oscillators, in which the damping force is velocity dependent and always opposes the motion of the oscillator. One way of producing this type of force is to use a piece of metal, such as aluminum, that moves through a nonuniform magnetic field. Explain why this technique is capable of producing a damping force.

An 8 -turn coil has square loops measuring \(0.200 \mathrm{~m}\) along a side and a resistance of \(3.00 \Omega\). It is placed in a magnetic field that makes an angle of \(40.0^{\circ}\) with the plane of each loop. The magnitude of this field varies with time according to \(B=1.50 t^{3}\), where \(t\) is measured in seconds and \(B\) in teslas. What is the induced current in the coil at \(t=2.00 \mathrm{~s} ?\)

A respiration monitor has a flexible loop of copper wire, which wraps about the chest. As the wearer breathes, the radius of the loop of wire increases and decreases. When a person in the Earth's magnetic field (assume \(\left.0.426 \cdot 10^{-4} \mathrm{~T}\right)\) inhales, what is the average current in the loop, assuming that it has a resistance of \(30.0 \Omega\) and increases in radius from \(20.0 \mathrm{~cm}\) to \(25.0 \mathrm{~cm}\) over \(1.00 \mathrm{~s}\) ? Assume that the magnetic field is perpendicular to the plane of the loop.

A supersonic aircraft with a wingspan of \(10.0 \mathrm{~m}\) is flying over the north magnetic pole (in a magnetic field of magnitude 0.500 G perpendicular to the ground) at a speed of three times the speed of sound (Mach 3). What is the potential difference between the tips of the wings? Assume that the wings are made of aluminum.

You have a light bulb, a bar magnet, a spool of wire that you can cut into as many pieces as you want, and nothing else. How can you get the bulb to light up? a) You can't. The bulb needs electricity to light it, not magnetism. b) You cut a length of wire, connect the light bulb to the two ends of the wire, and pass the magnet through the loop that is formed. c) You cut two lengths of wire and connect the magnet and the bulb in series.
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