Problem 38
Question
A long, thin solenoid has 900 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s. What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?
Step-by-Step Solution
Verified Answer
At 0.500 cm: \(1.02 \times 10^{-3}\) V/m; at 1.00 cm: \(2.04 \times 10^{-3}\) V/m.
1Step 1: Understand the Problem
We have a solenoid with 900 turns/m and radius 2.50 cm. The current increases at 36.0 A/s. We need to find the induced electric field at two different radial distances (0.500 cm and 1.00 cm) from the solenoid axis.
2Step 2: Recall Faraday's Law for Induced Electric Field
The induced electric field due to a changing magnetic field inside a solenoid can be described using Faraday's Law as: \[ E(2\pi r) = \mu_0 n \frac{di}{dt} \pi r^2 \]where \(E\) is the induced electric field, \(r\) is the distance from the center, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(\frac{di}{dt}\) is the rate of change of current.
3Step 3: Calculations for the Point 0.500 cm from Axis
Substitute the given values into the equation:\[ \frac{di}{dt} = 36.0 \text{ A/s}, \quad n = 900 \text{ turns/m}, \quad r = 0.005 \text{ m} \]The equation becomes:\[ E (2\pi \cdot 0.005) = (4\pi \times 10^{-7}) \times 900 \times 36 \times \pi \times (0.005)^2 \]Solve for \(E\):\[ E = \frac{(4\pi \times 10^{-7}) \times 900 \times 36 \times \pi \times (0.005)^2}{2\pi \times 0.005} \approx 1.02 \times 10^{-3} \text{ V/m} \]
4Step 4: Calculations for the Point 1.00 cm from Axis
Use the same process with \(r = 0.010 \text{ m}\):\[ E (2\pi \cdot 0.010) = (4\pi \times 10^{-7}) \times 900 \times 36 \times \pi \times (0.010)^2 \]Solve for \(E\):\[ E = \frac{(4\pi \times 10^{-7}) \times 900 \times 36 \times \pi \times (0.010)^2}{2\pi \times 0.010} \approx 2.04 \times 10^{-3} \text{ V/m} \]
5Step 5: Conclusion
The magnitude of the induced electric field at 0.500 cm is approximately \(1.02 \times 10^{-3} \text{ V/m}\), and at 1.00 cm, it is approximately \(2.04 \times 10^{-3} \text{ V/m}\).
Key Concepts
Induced Electric FieldSolenoidRate of Change of CurrentMagnetic Field
Induced Electric Field
The concept of an induced electric field emerges from the interaction between changing magnetic fields and electrical conductors. According to Faraday's Law of electromagnetic induction, when the magnetic field around a conductor changes, an electric field is induced. This electric field, in turn, can drive a current around a closed loop or circuit. The strength of this electric field depends on several factors:
- The rate at which the magnetic field changes.
- The area enclosed by the loop where the field is applied.
Solenoid
A solenoid is a coil of wire designed to produce a magnetic field when an electric current passes through it. In physics, solenoids are often idealized as long coils where the length is much greater than the diameter. This allows for a relatively uniform magnetic field across most of the interior of the coil and a negligible field outside it.Here's what makes solenoids interesting:
- The magnetic field inside a solenoid is strong and nearly uniform, making it useful for various applications like electromagnets and inductors.
- The field strength (B) inside the solenoid can be calculated using the formula: B = \(\ \mu_0 n i \) where \(\ \mu_0 \) (permeability of free space), \(\ n \) is the number of turns per unit length, and \(\ i \) is the current.
Rate of Change of Current
The rate of change of current is a critical factor in determining the magnitude of an induced electric field. In a solenoid setup, it refers to how quickly the current flowing through the coils is changing over time. This change is essential because a changing current results in a varying magnetic field.
Why is it important?
- The induced electric field strength is directly proportional to the rate at which the current changes. Simply put, faster changes in current magnify the resultant electric field.
- This concept is especially significant when designing devices like inductors, which rely on changing currents to function effectively.
Magnetic Field
A magnetic field is a vector field that describes the magnetic influence of electric currents and magnetic materials. It's present around magnetic materials and moving charges and is fundamental to the workings of electric motors, generators, and magnetic storage systems. In a solenoid, the generated magnetic field lines form closed loops and are most concentrated inside the coil.
Key characteristics of magnetic fields:
- They have both magnitude and direction; the direction of the magnetic field is the direction in which the north end of a compass needle points.
- In a solenoid, the magnetic field is concentrated and strong inside the coil while almost negligible outside.
- The strength of the field within a solenoid can be influenced by the number of turns in the coil, the current passing through, and the presence of a core material.
Other exercises in this chapter
Problem 36
A metal ring 4.50 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. Th
View solution Problem 37
The magnetic field within a long, straight solenoid with a circular cross section and radius \(R\) is increasing at a rate of \(dB/dt\). (a) What is the rate of
View solution Problem 39
A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate \(di/dt\). The induced electric fi
View solution Problem 41
A long, straight solenoid with a cross-sectional area of 8.00 cm\(^2\) is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350
View solution