Problem 7
Question
A solenoid has \(100 \overline{0}\) turns of wire, is \(0.320 \mathrm{~m}\) long, and carries a current of \(5.00 \mathrm{~A}\). What is the magnetic field at the center of the solenoid? Assume that its length is long in comparison with its diameter.
Step-by-Step Solution
Verified Answer
The magnetic field at the center is approximately 0.0196 T.
1Step 1: Identify the Formula
The magnetic field inside a long solenoid can be calculated using the formula \( B = \mu_0 \frac{nI}{L} \), where \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \mathrm{~T \cdot m/A}) \), \( n \) is the number of turns, \( I \) is the current, and \( L \) is the length of the solenoid.
2Step 2: Substitute Values into Formula
Substitute the provided values into the formula: \( B = (4\pi \times 10^{-7} \mathrm{~T \cdot m/A}) \times \frac{1000 \times 5.00}{0.320} \).
3Step 3: Calculate the Magnetic Field
Perform the calculation: \( B = (4\pi \times 10^{-7}) \times \frac{5000}{0.320} \approx 0.0196 \mathrm{~T} \).
Key Concepts
Magnetic Field CalculationSolenoid CharacteristicsElectric Current EffectsPermeability of Free Space
Magnetic Field Calculation
Understanding how to calculate the magnetic field inside a solenoid is essential to grasping its broader applications in physics and engineering. The calculation hinges on a straightforward formula: - \[ B = \mu_0 \frac{nI}{L} \]- Here, the magnetic field \( B \) is determined by multiplying the permeability of free space \( \mu_0 \) by the ratio of the number of turns \( n \) multiplied by the current \( I \) to the length of the solenoid \( L \). This formula is often used for solenoids that are significantly longer than they are wide, allowing us to uniformly assess the magnetic field in the center. By substituting the exercise's specifics into this equation, students can directly see the effect of each parameter on the resultant magnetic field. This conversion from descriptive terms to numerical values illustrates the direct, proportional relationship between these factors, offering practical insight into the solenoid's operation.
Solenoid Characteristics
Solenoids are a staple in electromagnetism due to their simple but effective design, characterized mainly by their coil winding structure.
A solenoid typically consists of:
- Numerous turns of wire usually wound in a spiral.
- A cylindrical core that may be air or a ferromagnetic material.
- Functions when an electric current passes through the wire, creating a magnetic field inside the coil.
Electric Current Effects
Electric current is the driving force behind the magnetism in solenoids, as it induces the magnetic field when it flows through the coil. The following are its primary effects:-
- The magnitude of the current directly affects the strength of the magnetic field \( B \), with an increase in current leading to a stronger magnetic field.
- The direction of the current contributes to the orientation of the generated magnetic field, determinable by the right-hand rule.
Permeability of Free Space
Understanding the concept of permeability of free space \( \mu_0 \) is crucial for grasping how magnetic fields interact in a vacuum or air. This constant is a measure of the ability of a vacuum to support a magnetic field and is defined as \( 4\pi \times 10^{-7} \) T·m/A.- Despite being in a state with no matter, such as the vacuum of free space, the magnetic field can still exist and propagate. Permeability of free space is a fundamental constant in physics and significantly impacts magnetic field calculations. In the formula to calculate the magnetic field in a solenoid, \( \mu_0 \) helps determine the intrinsic magnetic potential of the solenoid's interior devoid of additional material effects. This constant allows for a clear understanding of magnetic behavior across different environments, offering a foundational concept for fields like electromagnetism and quantum physics.
Other exercises in this chapter
Problem 2
Find the magnetic field at \(0.500 \mathrm{~m}\) from a long wire carrying a current of \(7.50 \mathrm{~A}\).
View solution Problem 5
Find the magnetic field at \(0.350 \mathrm{~m}\) from a long wire carrying a current of \(3.00 \mathrm{~A}\).
View solution Problem 8
A solenoid has 3000 turns of wire and is \(0.350 \mathrm{~m}\) long. What current is required to produce a magnetic field of \(0.100 \mathrm{~T}\) at the center
View solution Problem 9
A small solenoid is \(0.150 \mathrm{~m}\) in length and \(0.0150 \mathrm{~m}\) in diameter, and has \(60 \overline{0}\) turns of wire. What current is required
View solution