Problem 28 - University Physics with Modern Physics

Question

A long, thin solenoid has 900 turns per meter and radius \(2.50 \mathrm{cm} .\) The current in the solenoid is increasing at a uniform rate of 60.0 \(\mathrm{A} / \mathrm{s}\) . What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 \(\mathrm{cm}\) from the axis of the solenoid; (b) 1.00 \(\mathrm{cm}\) from the axis of the solenoid?

Step-by-Step Solution

Answer

(a) \(E \approx 3.18 \times 10^{-3} \ \text{V/m}\); (b) \(E \approx 1.59 \times 10^{-3} \ \text{V/m}\).

Step 1: Recall Faraday's Law of Induction

Faraday's law of induction states that the electromotive force (emf) induced in a closed loop is equal to the negative rate of change of magnetic flux through the loop. For a solenoid, the induced electric field \( E \) for a circular path inside it can be expressed as:\[ E \cdot 2\pi r = -\frac{d\Phi_B}{dt} \]where \(r\) is the distance from the axis, and \(\Phi_B\) is the magnetic flux.

Step 2: Calculate the Magnetic Flux Change Rate

The magnetic flux \(\Phi_B\) inside a solenoid is given by \( \Phi_B = B \cdot A \), where \( B \) is the magnetic field inside the solenoid, calculated as \( B = \mu_0 n I \). Here, \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \ \mathrm{T \cdot m/A}) \), \( n \) is the number of turns per unit length, and \( I \) is the current. Therefore, the rate of change of flux \( \frac{d\Phi_B}{dt} \) is:\[ \frac{d\Phi_B}{dt} = \mu_0 n A \frac{dI}{dt} \]where \( A = \pi r^2 \) is the area.

Step 3: Substitute Values

For part (a), with \( r = 0.500 \ \text{cm} = 0.0050 \ \text{m} \), substitute the given values:* \( n = 900 \ \text{turns/m} \)* \( \frac{dI}{dt} = 60.0 \ \text{A/s} \)Thus, the induced electric field at \( r = 0.500 \ \text{cm} \) is:\[E \cdot 2\pi (0.0050) = -\mu_0 (900) \pi (0.025^2) (60.0) \]Solve for \( E \).

Step 4: Calculation for Part (a)

Substituting into the equation and solving gives:\[E \cdot 2\pi (0.0050) = -4\pi \times 10^{-7} \times 900 \times \pi \times (0.025^2) \times 60 \]Solve for \( E \):\[E = \frac{-4\pi \times 10^{-7} \times 900 \times \pi \times (0.025^2) \times 60}{2\pi \times 0.0050} \]This simplifies to \(E \approx 3.18 \times 10^{-3} \ \text{V/m} \).

Step 5: Repetition for Part (b)

Repeat the process for part (b) with \( r = 1.00 \ \text{cm} = 0.010 \ \text{m} \). Using the same formula,\[E \cdot 2\pi (0.010) = -\mu_0 (900) \pi (0.025^2) (60.0)\]and solving for \( E \) yields:\[E = \frac{-4\pi \times 10^{-7} \times 900 \times \pi \times (0.025^2) \times 60}{2\pi \times 0.010}\]This simplifies to \(E \approx 1.59 \times 10^{-3} \ \text{V/m} \).

Key Concepts

Induced Electric Field

When the magnetic field inside a solenoid changes, it generates an induced electric field around it, which can be understood through Faraday's Law of Induction.
Faraday's law states that the electromotive force (emf) induced in a loop is equal to the negative rate of change of magnetic flux through that loop.

• The induced electric field is perpendicular to the changing magnetic field.
• The direction follows Lenz's Law, meaning it opposes the change creating it.

In a solenoid, which is essentially a tightly coiled wire, the induced electric field plays a crucial role in balancing the change in magnetic flux when the current changes.
To determine the magnitude of the induced electric field at a particular radius inside the solenoid, one can use the formula: \[ E \cdot 2\pi r = -\frac{d\Phi_B}{dt} \]This relates the induced electric field to the rate of change of magnetic flux through a circle of radius \(r\) inside the solenoid. Here, \(E\) is the electric field strength and \(\Phi_B\) is the magnetic flux.

Solenoid

A solenoid is a coil of wire that generates a magnetic field when an electric current passes through it. This component is widely used to create uniform magnetic fields in various applications.

• It consists of numerous loops of wire, known as turns, wrapped in a helical fashion.
• The strength of the magnetic field inside is largely determined by the number of turns per meter, denoted as \(n\).

The design of a solenoid takes advantage of Ampère's circuital law, which tells us that the magnetic field inside an infinitely long solenoid is given by \(B = \mu_0 n I\), where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current flowing through it.
The solenoid, being a straightforward structure, forms the perfect setting for studying magnetic flux changes and the resultant electric fields, especially in a physics context like Faraday's law.

Magnetic Flux

Magnetic flux, denoted by \(\Phi_B\), is a measure of the amount of magnetic field passing through a given area, such as the cross-section of a solenoid.
The magnetic flux inside a solenoid is given by the formula:\[ \Phi_B = B \cdot A \]where \(B\) is the magnetic field strength in teslas, and \(A\) is the area through which the field lines pass. For a solenoid with a circular cross-section, the area can be represented as \(\pi r^2\).

• The magnetic flux depends on both the magnetic field strength and the area exposed to this field.
• A change in the magnetic flux over time is what leads to an induced electromotive force (emf) as described by Faraday's law.

When dealing with solenoids, understanding how the flux changes with variations in current is key to comprehending the induced electric fields generated inside them.

Rate of Change of Current

The rate of change of current, often denoted as \(\frac{dI}{dt}\), is a critical factor in determining the induced electric field inside a solenoid.
The faster the current changes, the greater the change in magnetic flux, leading to a stronger induced electric field.

• In the context of a solenoid, as the current increases or decreases, the magnetic field generated by the solenoid also increases or decreases.
• This change in magnetic field alters the magnetic flux, resulting in an induced emf as described by Faraday's Law.

For example, if current in a solenoid increases at a constant rate, like the 60.0 A/s in the exercise, we calculate the change in magnetic flux using:\[ \frac{d\Phi_B}{dt} = \mu_0 n A \frac{dI}{dt} \]Here, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, \(A\) is the cross-sectional area, and \(\frac{dI}{dt}\) is how quickly the current changes. This equation shows how the rapid change in current directly influences the strength of the induced electric field.