At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 V. If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

Verified step by step guidance

1. First, recall the formula for the self-induced emf in an inductor: \( \text{emf} = -L \frac{dI}{dt} \), where \( L \) is the inductance, \( \frac{dI}{dt} \) is the rate of change of current, and \( \text{emf} \) is the electromotive force.
2. Given that the self-induced emf is 0.0160 V and the rate of change of current is 0.0640 A/s, use the formula to solve for the inductance \( L \): \( L = \frac{\text{emf}}{\frac{dI}{dt}} \). Substitute the given values to find \( L \).
3. Next, use the relationship between magnetic flux \( \Phi \), inductance \( L \), and current \( I \) in a solenoid: \( L = N \frac{\Phi}{I} \), where \( N \) is the number of turns. Rearrange this formula to solve for the magnetic flux \( \Phi \): \( \Phi = \frac{L \cdot I}{N} \).
4. Substitute the values for \( L \) (calculated in the previous step), \( I = 0.720 \) A, and \( N = 400 \) turns into the formula to find the average magnetic flux \( \Phi \) through each turn.
5. Ensure all units are consistent and perform the calculation to find the average magnetic flux through each turn of the solenoid.

Key Concepts

Inductance and Self-Induced EMF

Inductance is a property of an electrical conductor, such as a solenoid, that causes it to oppose changes in current. The self-induced electromotive force (emf) is generated when the current through the inductor changes, according to Faraday's law of induction. The emf is proportional to the rate of change of current and the inductance of the coil, given by the formula emf = -L * (di/dt), where L is the inductance.

Magnetic Flux

Magnetic flux through a surface is the product of the magnetic field and the area it penetrates, perpendicular to the field. It is a measure of the quantity of magnetism, taking into account the strength and extent of a magnetic field. For a solenoid, the magnetic flux through each turn is given by the formula Φ = B * A, where B is the magnetic field and A is the cross-sectional area of the solenoid.

Solenoid and Its Magnetic Field

A solenoid is a coil of wire designed to create a magnetic field when an electric current passes through it. The magnetic field inside a long solenoid is uniform and parallel to the axis of the solenoid, and its strength is given by B = μ₀ * (N/L) * I, where μ₀ is the permeability of free space, N is the number of turns, L is the length of the solenoid, and I is the current. This field is crucial for calculating the magnetic flux through each turn.