Problem 83 - University Physics with Modern Physics

Question

An insulated wire with mass \(m=5.40 \times 10^{-5} \mathrm{kg}\) is bent into the shape of an inverted U such that the horizontal part has a length \(l=15.0 \mathrm{cm} .\) The bent ends of the wire are partially immersed in two pools of mercury, with 2.5 \(\mathrm{cm}\) of each end below the mercury's surface. The entire structure is in a region containing a uniform \(0.00650-\mathrm{T}\) magnetic field directed into the page (Fig. 27.71\()\) . An electrical connection from the mercury pools is made through the ends of the wires. The mercury pools are connected to a \(1.50-\mathrm{V}\) battery and a switch \(\mathrm{S}\) . When switch \(\mathrm{S}\) is closed, the wire jumps 35.0 \(\mathrm{cm}\) into the air, measured from its initial position. (a) Determine the speed \(v\) of the wire as it leaves the mercury. (b) Assuming that the current \(I\) through the wire was constant from the time the switch was closed until the wire left the mercury, determine \(I\) (c) Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire.

Step-by-Step Solution

Answer

(a) The speed is approximately 2.62 m/s. (b) The current is about 0.545 A. (c) The resistance of the wire is roughly 2.75 Ω.

Step 1: Calculate the force due to the magnetic field

The wire experiences a magnetic force when it carries a current in the magnetic field. Using the formula for the magnetic force, \( F = I \times L \times B \), where \( F \) is the force, \( I \) is the current, \( L \) is the length of the wire in the magnetic field, and \( B \) is the magnetic field strength. Here, \( L = 0.15 \mathrm{m} \) and \( B = 0.00650 \mathrm{T} \). We need to find \( I \) first, but let's express \( F \) in terms of \( I \) for later use.

Step 2: Determine the initial velocity as the wire leaves the mercury

Using the work-energy principle, the mechanical work done by the magnetic force equals the potential energy gained by the wire when it rises. This is given by \( F \times h = m \times g \times h \), where \( h = 0.35 \mathrm{m} \) is the jump height, \( m = 5.40 \times 10^{-5} \mathrm{kg} \), and \( g = 9.81 \mathrm{m/s^2} \). From this, we solve for \( F = m \times g = 5.40 \times 10^{-5} \times 9.81 \).

Step 3: Calculate the speed of the wire (a)

Once the force \( F \) is known, the kinetic energy \( rac{1}{2}mv^2 \) is equal to the work done by the magnetic force, \( F \times h \). Solve for \( v \) using the equation \( v = \sqrt{\frac{2mgh}{m}} \). Substitute the values to find \( v = \sqrt{2 \times 9.81 \times 0.35} \).

Step 4: Calculate the current in the wire (b)

Substitute \( F = I \times L \times B \) and solve for \( I \) using the equivalent form \( I = \frac{F}{L \times B} \). Use \( F = 5.40 \times 10^{-5} \times 9.81 \). Substitute known values \( L = 0.15 \mathrm{m} \) and \( B = 0.00650 \mathrm{T} \) to find \( I \).

Step 5: Determine the resistance of the wire (c)

Ohm's Law \( V = I \times R \) helps to find the resistance \( R \). Rearranging gives \( R = \frac{V}{I} \). Use the potential difference \( V = 1.50 \mathrm{V} \) from the battery and the determined current \( I \) to solve for \( R \).

Key Concepts

Work-Energy Principle

The work-energy principle is a fundamental concept in physics that relates the mechanical work done on an object to its energy change. It states that the work done by all forces acting on an object equals the change in its kinetic energy. This principle is derived from the law of conservation of energy and can be used to solve problems involving forces and motion.

In the given exercise, when a magnetic force acts on the wire to lift it into the air, it does mechanical work on the wire. As the wire rises, this work translates into gravitational potential energy. The height the wire reaches helps us calculate the work done, since work equals the potential energy (PE) gained. The formula is:

• Work done by the force = change in gravitational PE = force \( F \times \) height \( h \).
• Potential energy gained = \( m \times g \times h \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.

This energy change allows us to determine how fast the wire is moving as it leaves the mercury, using kinetic energy principles.

Ohm's Law

Ohm's Law is a fundamental principle in the study of electric circuits, describing the relationship between voltage, current, and resistance. The law is typically stated as \( V = I \times R \), where \( V \) is the voltage across the circuit, \( I \) is the current flowing through it, and \( R \) is the resistance within the circuit.

In this problem, we need to determine the resistance of the moving wire once the current through it is known. By rearranging Ohm's Law, we can solve for resistance \( R \) using the formula:

• \( R = \frac{V}{I} \)
• Here, \( V \) is the voltage from the battery (1.50 V), and \( I \) is the current calculated from the magnetic force equations.

Understanding Ohm's Law helps us connect the physical movement of the wire with the electrical properties of the circuit.

Magnetic Field

A magnetic field is a vector field produced by moving electric charges and magnetic materials. It exerts forces on other moving charges or magnetic materials within the field, described by magnetic field strength \( B \) in teslas (T).

In our scenario, a magnetic field is applied perpendicular to the wire, creating a situation in which the wire experiences a force as current flows through it. This force can be calculated using the equation:

• Magnetic Force \( F = I \times L \times B \)
• Where \( I \) is the current, \( L \) is the length of the wire exposed to the magnetic field, and \( B \) is the magnetic field strength.

Since the wire is within a field of strength 0.00650 T, understanding the magnetic field's role is crucial in determining the force, and subsequently the wire’s motion.

Current in Circuits

Current is the flow of electric charge through a conductor, usually measured in amperes (A). In a given electric circuit, the current flows when a potential difference (voltage) is applied across its ends, driving charges through resistive elements.

In this particular problem, the current is induced by connecting the mercury pools to a 1.50-V battery. When the circuit is closed, the battery supplies energy that moves electrons, creating a current that interacts with the magnetic field to generate a mechanical force.

To solve for the current in the wire, we use the rearranged magnetic force formula:

• \( I = \frac{F}{L \times B} \), where \( F \) is the magnetic force computed from the energy principles, \( L = 0.15 \) m is the wire length in the field, and \( B = 0.00650 \) T.

Understanding how current relates to both the electrical and mechanical aspects of this problem is key to mastering the concept fully.