Problem 49 - University Physics with Modern Physics

Question

A very long, straight solenoid with a crosssectional area of 2.00 cm\(^2\) is wound with 90.0 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to \(i(t) = (0.160 A/s^2)t^2\). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.20 A?

Step-by-Step Solution

Answer

14.25 µV.

Step 1: Determine the Magnetic Field inside the Solenoid

The magnetic field inside a long solenoid can be calculated using the formula \( B = \mu_0 n i(t) \), where \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \) is the permeability of free space, \( n \) is the number of turns per meter, and \( i(t) \) is the current at time \( t \). First, we convert the number of turns per centimeter to turns per meter: \( n = 90.0 \times 100 = 9000 \text{ turns/m} \). Substituting these into the formula, the magnetic field \( B(t) = \mu_0 \cdot 9000 \cdot (0.160t^2) \).

Step 2: Calculate when Current is 3.20 A

Next, solve for \( t \) when \( i(t) = 3.20 \text{ A} \). The current as a function of time is given as \( i(t) = 0.160t^2 \), which can be rearranged to solve for \( t \):\[ 3.20 = 0.160t^2 \]Solve for \( t \):\[ t^2 = \frac{3.20}{0.160} = 20 \]\[ t = \sqrt{20} = 4.47 \text{ s} \]

Step 3: Determine EMF Induced

The emf \( \epsilon \) induced in a loop is given by Faraday's law of electromagnetic induction: \( \epsilon = -N \frac{d\Phi}{dt} \), where \( N = 5 \) is the number of turns in the secondary winding, and \( \Phi = BA \) is the magnetic flux through one loop. The rate of change of magnetic flux is necessary:\[ \frac{d\Phi}{dt} = A \frac{dB}{dt} \]Plug in the area in meters squared \( A = 2.00 \times 10^{-4} \, \text{m}^2 \) and the rate of change of \( B \). Since \( B = \mu_0 n i(t) \), we find \( \frac{dB}{dt} = \mu_0 n \frac{di}{dt} \).The derivative \( \frac{di}{dt} = 0.160 (2t) \). At \( t = 4.47s \), \( \frac{di}{dt} = 0.160 \times 2 \times 4.47 \).\[ \frac{di}{dt} = 1.43 \text{ A/s} \]Calculate \( \frac{dB}{dt} = \mu_0 \cdot 9000 \cdot 1.43 \).Finally, the emf is:\[ \epsilon = -5 \cdot (2.00 \times 10^{-4}) \cdot \mu_0 \cdot 9000 \cdot 1.43 \]

Step 4: Calculate the Magnitude of EMF

Substitute in known values:\[ \epsilon = -5 \cdot (2.00 \times 10^{-4}) \cdot (4\pi \times 10^{-7}) \cdot 9000 \cdot 1.43 \]Calculate,\[ \epsilon = -5 \cdot 2.85 \times 10^{-6} \]\[ \epsilon = -1.425 \times 10^{-5} \text{ V} \]The magnitude of \( \epsilon \) is \( 1.425 \times 10^{-5} \text{ V} \) or 14.25 \( \mu\text{V} \).

Key Concepts

Solenoid

A solenoid is a long coil of wire that is helix shaped and wound in tightly packed turns. When an electric current flows through it, a magnetic field is generated, resembling the field of a bar magnet. Solenoids are immensely important in electronics and electromagnetism.

Key features of solenoids include:

• Coil Turns: The more turns a solenoid has, the stronger the magnetic field it can create.
• Current: The current running through the solenoid wire determines the strength of the magnetic field generated.
• Core: Often filled with an iron core to intensify the magnetic field produced by the current.

An ideal solenoid is considered to be infinitely long with uniform turns, where the generated magnetic field is strong and uniform inside, and negligible outside. Understanding these characteristics helps in solving problems related to electromagnetic induction, such as those involving induced emf in a secondary coil placed near a solenoid.

Magnetic Field

The magnetic field is a vector field that describes the magnetic influence on moving charges, electric currents, and magnetic materials. In the context of a solenoid, the magnetic field is particularly important.

The strength of the magnetic field inside a long solenoid is given by the formula:
\( B = \mu_0 n i \), where:

• \( B \) is the magnetic field strength.
• \( \mu_0 \) is the permeability of free space, a constant factor \( 4\pi \times 10^{-7} \text{ Tm/A} \).
• \( n \) is the number of turns per unit length of the solenoid (usually turns per meter).
• \( i \) is the current flowing through the solenoid.

This formula shows how the magnetic field inside a solenoid is directly proportional to both the number of turns per unit length and the current. Thus, by adjusting these parameters, we can control the field strength, which is crucial in the exercise for finding the induced emf in a secondary coil.

Faraday's Law

Faraday's law of electromagnetic induction is a fundamental principle that describes how changing magnetic fields can induce an electromotive force (emf) in a coil. It is expressed by the equation:
\( \epsilon = -N \frac{d\Phi}{dt} \), where:

• \( \epsilon \) is the induced emf.
• \( N \) is the number of turns in the coil where emf is induced.
• \( \Phi \) is the magnetic flux through one loop of the coil.

The negative sign in the equation is in accordance with Lenz's law, indicating that the induced emf will generate a current that opposes the change in magnetic flux. When applied to the problem of a solenoid and secondary winding, Faraday's law is essential for finding the magnitude of the emf, as it relates to the changing magnetic field created by the solenoid's current, and how this change influences the secondary coil.

Magnetic Flux

Magnetic flux is a measure of the quantity of magnetism, taking into account the strength and extent of a magnetic field. It is represented by the symbol \( \Phi \) and is mathematically defined as:
\( \Phi = B \times A \times \cos(\theta) \), where:

• \( B \) is the magnetic field strength.
• \( A \) is the area through which the field lines pass.
• \( \theta \) is the angle between the field lines and the normal (perpendicular) to the surface.

In most solenoid problems, we assume \( \theta = 0 \), making \( \cos(\theta) = 1 \), thus simplifying the flux to \( \Phi = B \times A \).

Understanding magnetic flux is vital to solving problems related to electromagnetic induction as it connects the change in the solenoid's magnetic field to the induced emf in another coil via Faraday's law. By knowing how to calculate magnetic flux and its rate of change, you can determine how strong an emf is induced.