/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A horizontal wire is hung from t... [FREE SOLUTION] | 91Ó°ÊÓ

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A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.20 \(\mathrm{m}\) and a mass of 0.080 kg. A uniform magnetic field of magnitude 0.070 \(\mathrm{T}\) is directed from the ceiling to the floor.When a current of \(I=42 \mathrm{A}\) exists in the wire, the wire swings upward and, at equilibrium, makes an angle \(\phi\) with respect to the vertical, as the drawing shows. Find (a) the angle \(\phi\) and \((b)\) the tension in each of the two strings.

Short Answer

Expert verified
(a) \( \phi \approx 36.87^\circ \); (b) Tension in each string: \( 0.98 \ \mathrm{N} \).

Step by step solution

01

Understanding the problem

We have a wire that swings upward due to a magnetic field when a current passes through it. We need to find the angle, \( \phi \), the wire makes with the vertical and the tension in the strings.
02

Applying the magnetic force formula

The magnetic force acting on the wire is given by \( F_B = I \cdot L \cdot B \cdot \sin(\theta) \). Here, \( I = 42 \ \mathrm{A} \), \( L = 0.20 \ \mathrm{m} \), and \( B = 0.070 \ \mathrm{T} \). Since the force is perpendicular to both the current and the magnetic field, \( \theta = 90^\circ \), making \( \sin(\theta) = 1 \).
03

Calculate magnetic force

Substitute the values into the formula: \( F_B = 42 \times 0.20 \times 0.070 \times 1 = 0.588 \ \mathrm{N} \). This is the force that causes the wire to swing.
04

Analyzing forces to find angle \(\phi\)

At equilibrium, the magnetic force \( F_B \), gravitational force \( mg \), and tension \( T \) must balance. The vertical component of the tension is \( T \cos(\phi) = mg \), and the horizontal component is \( T \sin(\phi) = F_B \).
05

Solve for angle \( \phi \)

From \( \frac{T \sin(\phi)}{T \cos(\phi)} = \frac{F_B}{mg} \), we have \( \tan(\phi) = \frac{F_B}{mg} \). Calculate \( mg = 0.080 \times 9.8 = 0.784 \ \mathrm{N} \). Then, \( \phi = \tan^{-1}\left(\frac{0.588}{0.784}\right) \approx 36.87^\circ \).
06

Find the tension in the strings

Using \( T \cos(\phi) = mg \), solve for \( T \). We have \( T = \frac{mg}{\cos(\phi)} = \frac{0.784}{\cos(36.87^\circ)} \approx 0.98 \ \mathrm{N} \). Each string has the same tension, so the tension in each is \( 0.98 \ \mathrm{N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetostatics
Magnetostatics is the study of magnetic fields in systems where the currents are steady and unchanging over time. This concept is crucial for understanding how magnetic forces originate and how they interact with materials such as current-carrying wires. In our exercise, a magnetic field is directed from the ceiling to the floor and affects a wire with a steady current passing through it.
- Magnetic fields are generated by electric currents, according to the Biot-Savart Law, which implies that the wire carrying a direct current will create a magnetic field. - The magnetic field interacts with the wire, producing a force as described by the Lorentz force law, which states the force on a current-carrying conductor is the product of current (I), length of wire (L), magnetic field (B), and the sine of the angle between the current and the magnetic field. - In our specific case, the angle is 90 degrees because the magnetic field is perpendicular to the current, leading to maximum force acting on the wire. Understanding how magnetic fields interact with currents is a key part of magnetostatics, allowing us to analyze and predict the forces a wire will experience.
Equilibrium in physics
Equilibrium occurs when all forces acting on an object cancel each other out, leading to a situation where there is no net force causing further motion or rotation. This is the scenario in our exercise when the wire comes to rest at an angle, \( \phi \), with respect to the vertical.
- For the wire to be in equilibrium, the sum of forces in both horizontal and vertical directions must be zero.- The vertical forces include the gravitational force pulling down and the vertical component of the tension pulling upwards.- The horizontal component of the tension must balance the horizontal magnetic force acting on the wire.- By resolving these forces using trigonometry, one can establish a balance of forces equations: \( T \cos(\phi) = mg \) and \( T \sin(\phi) = F_B \), which derive directly from the condition of equilibrium.These conditions allow us to solve for the angle and tension, ensuring that the wire is steady and not moving.
Trigonometry in physics
Trigonometry is an essential tool in physics for breaking forces and motion into their components. In this exercise, we use it to resolve the tension in the strings into vertical and horizontal components.
- The angle \( \phi \) the wire makes with the vertical allows us to use trigonometric functions such as sine and cosine to split the tension force into parts.- \(T \cos(\phi)\) represents the vertical component that counters gravity, while \(T \sin(\phi)\) represents the horizontal component that counters the magnetic force.This approach is vital for solving problems involving forces acting at angles and enables the calculation of unknown variables in equilibrium situations.
Current-carrying wire in a magnetic field
When a wire carrying an electric current is placed in a magnetic field, it experiences a force. This fundamental principle lies at the heart of many electromagnetic applications.
- The force is known as the magnetic force or Lorentz force, and it is calculated using the formula: \( F_B = I \cdot L \cdot B \cdot \sin(\theta) \).- In our exercise, since the wire is horizontal and the magnetic field is vertical, \( \theta = 90^\circ \), meaning the sine of the angle is 1, leading to the maximum possible magnetic force.- This force causes the wire to swing upwards until it reaches equilibrium, showing how magnetic fields can cause mechanical motion in conductors.Understanding this interaction is vital in studying electromagnetism, giving insight into the design of electric motors, generators, and other technologies reliant on magnetic forces.

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Most popular questions from this chapter

A long solenoid has a length of 0.65 \(\mathrm{m}\) and contains 1400 turns of wire. There is a current of 4.7 \(\mathrm{A}\) in the wire. What is the magnitude of the magnetic field within the solenoid?

Two charged particles move in the same direction with respect to the same magnetic field. Particle 1 travels three times faster than particle 2. However, each particle experiences a magnetic force of the same magnitude. Find the ratio \(\left|q_{1}\right| /\left|q_{2}\right|\) of the magnitudes of the charges.

Particle 1 and particle 2 have masses of \(m_{1}=2.3 \times 10^{-8} \mathrm{kg}\) and \(m_{2}=5.9 \times 10^{-8} \mathrm{kg},\) but they carry the same charge \(q .\) The two particles accelerate from rest through the same electric potential difference \(V\) and enter the same magnetic field, which has a magnitude \(B\). The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is \(r_{1}=12 \mathrm{cm} .\) What is the radius (in cm) of the circular path for particle 2\(?\)

A square coil and a rectangular coil are each made from the same length of wire. Each contains a single turn. The long sides of the rectangle are twice as long as the short sides. Find the ratio \(\tau_{\text { square }} / \tau_{\text { rectangle }}\) of the maximum torques that these coils experience in the same magnetic field when they contain the same current.

A small compass is held horizontally, the center of its needle a distance of 0.280 \(\mathrm{m}\) directly north of a long wire that is perpendicular to the earth's surface. When there is no current in the wire, the compass needle points due north, which is the direction of the horizontal component of the earth's magnetic field at that location. This component is parallel to the carth's surface. When the current in the wire is \(25.0 \mathrm{A},\) the needle points \(23.0^{\circ}\) east of north. (a) Does the current in the wire flow toward or away from the earth’s surface? (b) What is the magnitude of the horizontal component of the earth’s magnetic field at the location of the compass?

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