Question

A 100 -turn solenoid of length \(8.00 \mathrm{~cm}\) and radius \(6.00 \mathrm{~mm}\) carries a current of 0.400 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

Answer: There is no change in the energy stored in the magnetic field inside the solenoid when the current is reversed. The change in energy is 0 J.

Step-by-step solution

Determine magnetic field inside the solenoid

The formula for the magnetic field inside a solenoid is given by: \(B=\mu_0 \cdot n \cdot I\), where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} T\cdot m/A\)), \(n\) is the number of turns per unit length, and \(I\) is the current flowing through the solenoid. We have a solenoid with 100 turns and a length of 8 cm (0.08 m), so we can calculate the number of turns per unit length: \(n = \frac{100}{0.08} = 1250~ turns/m\) When the current is 0.400 A from right to left, the magnetic field \(B_1\) is given by: \(B_1 = \mu_0 \cdot n \cdot I_1\) When the current is 0.400 A from left to right, the magnetic field \(B_2\) is the same but with the opposite direction. Therefore, \(B_2 = -\mu_0 \cdot n \cdot I_1\) Now, calculate the magnetic fields \(B_1\) and \(B_2\).

Compute the magnetic field for both directions

Compute the magnetic field \(B_1\) and \(B_2\) \(B_1 = 4\pi \times 10^{-7} T\cdot m/A \cdot 1250~ turns/m \cdot 0.400 A = 0.0628 T\) \(B_2 = - 0.0628 T\)

Compute the energies stored in the magnetic field for both directions

The formula for the energy density \(u\) inside the solenoid is given by: \(u = \frac{B^2}{2\mu_0}\) The volume of the solenoid is given by: \(V = \pi \cdot (\text{radius})^2 \cdot \text{length} = \pi \cdot (0.006 m)^2 \cdot 0.08 m\) Now we can compute the energy stored in the magnetic field for both directions: \(E_{B1}\) and \(E_{B2}\): \(E_{B1} = u_{B1} \cdot V = \frac{B_1^2}{2\mu_0} \cdot V\) \(E_{B2} = u_{B2} \cdot V = \frac{B_2^2}{2\mu_0} \cdot V\)

Calculate the energies stored in the magnetic field for both directions

Compute the energies \(E_{B1}\) and \(E_{B2}\) \(E_{B1} = \frac{(0.0628 T)^2}{2\cdot 4\pi \times 10^{-7} T\cdot m/A} \cdot \pi \cdot (0.006 m)^2 \cdot 0.08 m = 2.658 \times 10^{-4} J\) \(E_{B2} = \frac{(-0.0628 T)^2}{2\cdot 4\pi \times 10^{-7} T\cdot m/A} \cdot \pi \cdot (0.006 m)^2 \cdot 0.08 m = 2.658 \times 10^{-4} J\) We can see that the energies \(E_{B1}\) and \(E_{B2}\) are the same, even though the direction of the magnetic field has changed.

Find the change in energy stored in the magnetic field inside the solenoid

Since the energies stored in the magnetic field for both directions, \(E_{B1}\) and \(E_{B2}\), are the same, the change in energy is zero. ΔE = \(E_{B2} - E_{B1} = 0 J\) Hence, there is no change in the energy stored in the magnetic field inside the solenoid when the current is reversed.

Key concepts

Magnetic Field Inside Solenoid

Understanding the magnetic field generated within a solenoid is crucial when studying electromagnetism. A solenoid is essentially a coil of wire, and when an electric current flows through it, a magnetic field is created in and around the coil. The strength of this magnetic field inside the solenoid is determined by several factors, including the number of turns of the wire, the length of the solenoid, and the current passing through it.

The formula widely used to express this relationship is \( B = \mu_0 \cdot n \cdot I \), where \( B \) represents the magnetic field, \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} T\cdot m/A\)), \( n \) is the number of turns per unit length of the solenoid, and \( I \) signifies the current. It's worth noting that the formula assumes a uniform magnetic field and tightly wound coils, which create an almost ideal scenario for the field inside the solenoid.

To deepen student comprehension, it's beneficial to explain how solenoids are used in real-world applications, like electromagnetic door locks and electromagnets for lifting scrap metal, where controlling the current allows for the manipulation of the magnetic field strength.

Energy Density of Magnetic Field

Moving further into the realm of solenoid analysis, we encounter the concept of energy density of a magnetic field. This represents the energy stored in a magnetic field per unit volume. For a solenoid's magnetic field, the energy density \( u \) is described by the equation \( u = \frac{B^2}{2\mu_0} \), where \( B \) is the magnetic field and \( \mu_0 \) is the permeability of free space.

Despite its abstract nature, understanding this energy density is important for applications such as superconducting magnets used in MRI machines, where the energy stored is directly related to the magnetic field's strength. This concept also highlights that the energy density, and thus the stored energy, exclusively depends on the magnitude of the magnetic field and not its direction. Such knowledge reinforces why reversing the current direction in our solenoid example doesn't change the amount of energy stored.

Permeability of Free Space

The permeability of free space, denoted as \( \mu_0 \), is a fundamental constant that plays a key role in electromagnetism. It essentially characterizes the ability of a vacuum to support the formation of a magnetic field. The value of \( \mu_0 \) is precisely defined as \(4\pi \times 10^{-7} T\cdot m/A\), or Tesla meter per Ampere.

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