Problem 14 - University Physics with Modern Physics

Question

A long, straight solenoid has 800 turns. When the current in the solenoid is \(2.90 \mathrm{A},\) the average flux through each turn of the solenoid is \(3.25 \times 10^{-3} \mathrm{Wb}\) . What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 7.50 \(\mathrm{mV} ?\)

Step-by-Step Solution

Answer

The rate of change of current is \(8.36 \times 10^{-3}\) A/s.

Step 1: Understanding the Problem

We need to find the rate of change of the current in a solenoid, given some initial conditions, so that the self-induced emf is 7.50 mV. We know the number of turns, the current, and the magnetic flux.

Step 2: Using the Formula for Self-Induced EMF

Self-induced emf in a solenoid is given by the formula:\[\mathcal{E} = -L \frac{di}{dt}\]where \(\mathcal{E}\) is the emf, \(L\) is the inductance, and \(\frac{di}{dt}\) is the rate of change of current. We need to express \(\frac{di}{dt}\) in terms of known quantities.

Step 3: Calculating Inductance of Solenoid

Inductance \(L\) of a solenoid is given by:\[L = \frac{N \Phi}{I}\]where \(N\) is the number of turns, \(\Phi\) is the magnetic flux per turn, and \(I\) is the current. Plugging in given values: \(N = 800\), \(\Phi = 3.25 \times 10^{-3}\) Wb, \(I = 2.90\) A.

Step 4: Substituting Values into Inductance Formula

Calculate \(L\):\[L = \frac{800 \times 3.25 \times 10^{-3}}{2.90} = 0.8966 \text{ H}\]

Step 5: Rearranging EMF Formula for Rate of Change of Current

We need to find \(\frac{di}{dt}\):\[7.50 \times 10^{-3} = -0.8966 \cdot \frac{di}{dt}\]Solving for \(\frac{di}{dt}\):\[\frac{di}{dt} = \frac{-7.50 \times 10^{-3}}{-0.8966} = 8.36 \times 10^{-3} \text{ A/s}\]

Step 6: Final Answer

The magnitude of the rate of change of the current \(\left|\frac{di}{dt}\right|\) must be \(8.36 \times 10^{-3}\) A/s.

Key Concepts

Solenoid

A solenoid is a long coil of wire, often wrapped in a cylindrical shape. When electric current passes through it, the solenoid generates a magnetic field similar to that of a bar magnet. The strength of this magnetic field is directly proportional to the number of turns in the solenoid and the current flowing through it. Solenoids are widely used because they can create controlled magnetic fields of varying strength by adjusting the current. In practical applications, you will find solenoids in devices like electromagnetic locks and valves.

• A solenoid becomes stronger with more turns or increased current.
• It forms the fundamental structure for electromagnets.
• Is utilized in various electronic applications.

Self-Induced EMF

Self-induced electromotive force (emf) occurs when a change in current through a coil induces a voltage (emf) in the same coil. This phenomenon primarily results from a changing magnetic field produced by the current change in the coil.The induced emf opposes the change in current, according to Lenz's law. Mathematically, it is represented by \[ \mathcal{E} = -L \frac{di}{dt} \]where \( \mathcal{E} \) is the emf, \( L \) is the inductance, and \( \frac{di}{dt} \) is the rate of change of current. This self-induction property is crucial for understanding how circuits and electromagnetic devices operate.

• Causes a voltage to oppose current change.
• Fundamental in alternating current (AC) circuits.
• Makes transformers and inductors possible.

Magnetic Flux

Magnetic flux captures the amount of magnetic field passing through a given area. It is an essential concept for understanding electromagnetic principles. In the context of a solenoid, it refers to the magnetic field passing through each turn of the coil.The formula of magnetic flux \( \Phi \) is expressed as:\[ \Phi = B \cdot A \]where \( B \) is the magnetic field strength and \( A \) is the area it penetrates.

• Measured in Webers (Wb).
• Indicates field strength and area coverage.
• Vital for calculating induced currents and voltages.

Inductance

Inductance is the property of a coil that quantifies its ability to induce an emf as a result of changes in current. This effect results from the magnetic field interactions within the coil.The inductance \( L \) of a solenoid is given by:\[ L = \frac{N \Phi}{I} \]where \( N \) is the number of turns, \( \Phi \) is the average magnetic flux through each turn, and \( I \) is the current.With this relationship, inductance becomes a measure of a solenoid's capability to oppose changes in current. It is crucial for the operation of many electrical devices such as transformers and inductors.

• Measured in Henry (H).
• Determines the magnetic energy storage potential.
• Essential in filters and tuning circuits.