Question

A long solenoid with length \(3.0 \mathrm{~m}\) and \(n=290\) turns \(/ \mathrm{m}\) carries a current of \(3.0 \mathrm{~A} .\) It stores \(2.8 \mathrm{~J}\) of energy. What is the cross-sectional area of the solenoid?

Short answer

Answer: The cross-sectional area of the solenoid is approximately \(1.02 \times 10^{-3} \mathrm{~m^2}\).

Step-by-step solution

Calculate magnetic field inside solenoid

First, we need to find the magnetic field inside the solenoid (B). Use the formula for magnetic field inside a long solenoid: $$ B = \mu_0 n I $$ Plug in the values for \(\mu_0 = 4\pi × 10^{-7} \mathrm{~T m/A}\), n = 290 turns/m, and I = 3.0 A: $$ B = (4\pi × 10^{-7} \mathrm{~T m/A}) \cdot (290 \mathrm{~turns/m}) \cdot (3.0 \mathrm{~A}) $$ Calculate the magnetic field B: $$ B \approx 3.63 \times 10^{-3} \mathrm{~T} $$

Find volume of the solenoid

Use the formula for stored energy inside a long solenoid: $$ W = \frac{1}{2} \cdot \frac{B^2}{\mu_0} \cdot V $$ We know W = 2.8 J, B = 3.63 × 10^{-3} T, and \(\mu_0 = 4\pi × 10^{-7} \mathrm{~T m/A}\). Plug in the values and solve for volume V: $$ 2.8 \mathrm{~J} = \frac{1}{2} \cdot \frac{(3.63 \times 10^{-3} \mathrm{~T})^2}{4\pi × 10^{-7} \mathrm{~T m/A}} \cdot V $$ Calculate the volume V: $$ V \approx 3.07 \times 10^{-3} \mathrm{~m^3} $$

Find the cross-sectional area of the solenoid

We have the volume V and the length L of the solenoid. Use the relationship between volume, length, and cross-sectional area: $$ V = A \cdot L $$ We know V = 3.07 × 10^{-3} m³ and L = 3.0 m. Plug in the values and solve for cross-sectional area (A): $$ 3.07 \times 10^{-3} \mathrm{~m^3} = A \cdot (3.0 \mathrm{~m}) $$ Calculate the cross-sectional area A: $$ A \approx 1.02 \times 10^{-3} \mathrm{~m^2} $$ The cross-sectional area of the solenoid is approximately \(1.02 \times 10^{-3} \mathrm{~m^2}\).

Key concepts

Electromagnetism

Electromagnetism is a fascinating field of physics that deals with the forces caused by electric and magnetic fields. It beautifully unites electricity and magnetism under one fundamental principle. One of its real-world applications is within devices known as solenoids. Solenoids are coil-shaped wires that convert electrical energy into magnetic energy, enabling them to act like magnets. This conversion is crucial in engineering and technology, including in devices like electromechanical valves and MRI machines. When a current flows through the solenoid, it creates a uniform magnetic field, which is strongest inside the coil. This property is harnessed in various applications, like electromagnetic locks or inductors. For students delving into this topic, understanding how electric current, coil turns, and the solenoid's form factor affect the magnetic field is essential. These factors differentiate a solenoid from a regular wire in creating a significant magnetic field.

Magnetic Field Calculation

Calculating the magnetic field of a solenoid involves understanding how factors like the current and the coil's turns impact the magnetic strength. The formula to find the magnetic field inside a solenoid is given by: \( B = \mu_0 n I \). Here, \( B \) represents the magnetic field, \( \mu_0 \) is the permeability of free space (approximately \( 4\pi \times 10^{-7} \mathrm{~T\cdot m/A} \)), \( n \) is the number of turns per meter, and \( I \) signifies the current in amperes. This formula shows how an increased number of turns or a stronger current can enhance the magnetic field inside the solenoid. By substituting known values into the equation, one can calculate the exact field strength, helping to predict how the solenoid will behave in a circuit. Solving problems with this equation requires careful attention to detail in units and the precision of constants, allowing students to grasp the relationship between electricity and magnetism in a quantitative manner.

Energy Storage in Solenoids

Energy storage is a critical function of solenoids, leveraged in both simple circuits and sophisticated applications like magnetic resonance imaging. When a solenoid stores energy, it does so in its magnetic field. The energy stored, \( W \), can be calculated using the formula: \( W = \frac{1}{2} \cdot \frac{B^2}{\mu_0} \cdot V \). In this equation, \( B \) stands for the magnetic field, \( \mu_0 \) is the permeability of free space, and \( V \) represents the volume of the solenoid. By rearranging this formula, students can find unknown values like the solenoid's volume or cross-sectional area when other parameters are known. This highlights the capacity of solenoids to sustain energy, which is necessary for applications requiring steady magnetic fields. This energy capacity is essential for consistent operation in devices like transformers and inductors, demonstrating the solenoid's role in energy-efficient technological design.