Question

A copper wire with density \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\) is formed into a circular loop of radius \(50.0 \mathrm{~cm} .\) The cross-sectional area of the wire is \(1.00 \cdot 10^{-5} \mathrm{~m}^{2},\) and a potential difference of \(0.012 \mathrm{~V}\) is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) ? The loop rotates about an axis through a diameter.

Short answer

Based on the given information of a copper wire loop with density, radius, cross-sectional area, and potential difference placed in a magnetic field, calculate the maximum angular acceleration. Answer: The maximum angular acceleration of the copper wire loop when placed in a magnetic field of magnitude 0.25 T is 0.1001 rad/s².

Step-by-step solution

Calculate the mass of the wire

To calculate the mass of the wire, we first need to find its length. From the geometry of the circular loop, we know that the length is equal to the circumference of the circle. The formula for the circumference is \(C = 2 \pi r\), where \(r\) is the radius. \(C = 2 \pi (50.0 \times 10^{-2}) \mathrm{~m} = \pi \mathrm{~m}\) Now, we can find the volume of the wire, which is the product of its cross-sectional area and its length. \(V = A \times C = (1.00 \times 10^{-5} \mathrm{~m}^2) \times (\pi \mathrm{~m}) = \pi \times 10^{-5} \mathrm{~m}^3\) Next, we use the density, \(\rho\), to find the mass of the wire. \(m = \rho \times V = 8960 \mathrm{~kg/m}^3 \times (\pi \times 10^{-5} \mathrm{~m}^3) = 28.274 \mathrm{~kg}\)

Find the current flowing through the wire

To find the current flowing through the wire, we'll use Ohm's Law, \(V = IR\), where \(V\) is the potential difference, \(I\) is the current, and \(R\) is the resistance. We can find the resistance by rearranging the formula, \(R = \frac{V}{I}\). We need to find the resistivity of copper, which is given as \(\rho_c = 1.68 \times 10^{-8} \Omega \cdot \mathrm{m}\) Using the formula for resistance in a cylindrical wire, \(R = \frac{\rho_c L}{A}\), we can find the resistance of the wire loop: \(R = \frac{1.68 \times 10^{-8} \Omega \cdot \mathrm{m} \times \pi \mathrm{~m} }{1.00 \times 10^{-5} \mathrm{~m}^2} = 5.305 \times 10^{-3} \Omega\) Now, we can calculate the current flowing through the wire: \(I = \frac{V}{R} = \frac{0.012 \mathrm{~V}}{5.305 \times 10^{-3} \Omega} = 2.26 \mathrm{~A}\)

Calculate the force exerted by the magnetic field

The magnetic force on the loop can be calculated using the formula: \(F = B I L\), where \(B\) is the magnetic field, \(I\) is the current, and \(L\) is the length of the wire acting perpendicular to the magnetic field. Since the wire loop is circular and placed in a uniform magnetic field, half of the loop experiences a force in one direction, and the other half experiences a force in the opposite direction. Thus, the net force on the loop is zero, but there is a net torque that is exerted on the loop. We can take the force acting on one half of the loop and multiply it by the radius to find the torque.

Calculate the torque exerted by the magnetic field

Torque can be calculated as follows: \(\tau = r \times F = r \times B I L\) Since only half of the loop is considered, and the other half is contributing torque in the opposite direction, we need to divide the length of the wire (\(L\)) by 2. Hence, \(\tau = (50.0 \times 10^{-2})\times (0.25 \mathrm{T}) \times (2.26\mathrm{A}) \times (\frac{\pi}{2} \mathrm{m}) = 1.77\mathrm{ N\cdot m}\)

Calculate the mass moment of inertia

To find the mass moment of inertia for the wire loop (circle) rotating about an axis through a diameter, we can use the perpendicular axis theorem. The mass moment of inertia of the wire loop about the axis is given by: \(I = \frac{1}{2} m r^2\) \(I = \frac{1}{2}(28.274 \mathrm{~kg})(50.0 \times 10^{-2} \mathrm{~m})^2 = 17.67 \mathrm{~kg \cdot m^2}\)

Calculate the maximum angular acceleration

Finally, we can find the angular acceleration using the equation for Newton's second law in rotational form: \(\tau = I \alpha\) We can rearrange the equation to find the angular acceleration: \(\alpha = \frac{\tau}{I} = \frac{1.77\mathrm{ N\cdot m}}{17.67 \mathrm{~kg \cdot m^2}} = 0.1001 \mathrm{~rad/s^2}\) So, the maximum angular acceleration of the loop when placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) is \(0.1001 \mathrm{~rad/s^2}\).

Key concepts

Magnetic Torque on Current Loop

When a current-carrying loop is placed within a magnetic field, a magnetic torque is generated due to the interaction of the magnetic field with the current. This torque causes the loop to rotate around an axis that is perpendicular to the magnetic field lines. The magnitude of the magnetic torque is given by the formula \( \tau = NIAB \sin(\theta) \), where \( N \) represents the number of turns in the coil, \( I \) is the current flowing through the coil, \( A \) is the area of the loop, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the plane of the coil and the magnetic field lines.

In the context of the exercise, the torque on our single-loop copper wire can be simplified to \( \tau = r \times B \times I \times L/2 \), considering the loop is placed perpendicular to the uniform magnetic field so that \( \sin(\theta) = 1 \). This formula helps us understand how various physical quantities such as current, magnetic field strength, and the dimensions of the loop, specifically the radius and length of the wire affect the torque exerted on the current loop.

Mass Moment of Inertia

The mass moment of inertia is a measure of an object's resistance to change in its rotational motion about a given axis. It is the rotational equivalent of mass in linear motion. The moment of inertia depends on the mass of the object and how that mass is distributed with respect to the axis of rotation. For a thin circular loop of radius \( r \) and mass \( m \) rotating around an axis through its diameter, the mass moment of inertia can be determined by the formula \( I = \frac{1}{2}mr^2 \).

This value is crucial when calculating the angular acceleration of the loop in response to an applied torque, as represented in the exercise. Once the mass moment of inertia is known, applying Newton's second law for rotation, \( \tau = I\alpha \), allows us to solve for the angular acceleration \( \alpha \). Importantly, the mass moment of inertia is unique to the geometry and the axis of rotation of the object, as different shapes and axes yield different formulas.

Ohm's Law

Ohm's Law is a fundamental principle in the field of electromagnetism that describes the relationship between voltage, current, and resistance within an electrical circuit. The law is usually expressed by the equation \( V = IR \), where \( V \) stands for voltage (the potential difference), \( I \) for the electric current, and \( R \) for the resistance. Ohm's Law indicates that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant.

In our exercise, Ohm's Law is used to calculate the current flowing through the copper wire loop when a known potential difference is applied. Understanding this law is essential for anyone working with electrical circuits as it enables the prediction and control of current flow. It also lays the foundation for more advanced concepts in electrical engineering and physics.

Magnetic Force

A magnetic force arises when a charged particle moves through a magnetic field. The force is perpendicular to both the direction of the particle's movement and the magnetic field lines. This magnetic force does not work on the particle, which means it doesn't change the particle's kinetic energy, but it can change the direction of the particle's path. For a current-carrying wire, the magnetic force can be calculated using the formula \( F = BIL \sin(\theta) \), with \( B \) being the magnetic field strength, \( I \) the current, \( L \) the length of wire within the magnetic field, and \( \theta \) the angle between the current and the magnetic field.

In our exercise, even though the magnetic force on the entire loop is zero due to equal and opposite forces on the two halves of the loop, this force is responsible for creating the magnetic torque that leads to the loop's rotation. The concept of magnetic force is vital not only in academic problems but also in real-world applications, such as electric motors, generators, and magnetic resonance imaging (MRI) machines.