/*! 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 43 The \(1200-\) -turn coil in a dc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The \(1200-\) -turn coil in a dc motor has an area per turn of \(1.1 \times 10^{-2} \mathrm{m}^{2}\) The design for the motor specifies that the magnitude of the maximum torque is 5.8 \(\mathrm{N} \cdot \mathrm{m}\) when the coil is placed in a \(0.20-\mathrm{T}\) magnetic field What is the current in the coil?

Short Answer

Expert verified
The current in the coil is approximately 2.20 A.

Step by step solution

01

Understand the Given Problem

We have a coil with 1200 turns in a DC motor. Each turn has an area of \(1.1 \times 10^{-2} \mathrm{m}^2\). The maximum torque produced by this coil in a \( 0.20 \mathrm{T} \) magnetic field is \(5.8 \mathrm{N} \cdot \mathrm{m}\). We need to find the current in the coil.
02

Write Down the Torque Formula for a Coil

The formula for the torque (\(\tau\)) on a coil in a magnetic field is \( \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \), where \(n\) is the number of turns, \(I\) is the current, \(A\) is the area of one turn, \(B\) is the magnetic field, and \(\theta\) is the angle between the field and the normal to the coil. For maximum torque, \(\sin(\theta)\) equals 1, since \(\theta = 90^\circ\).
03

Substitute Given Values into the Formula

Substitute \( n = 1200 \), \( A = 1.1 \times 10^{-2} \mathrm{m}^2 \), \( B = 0.20 \mathrm{T} \), and \(\tau = 5.8 \mathrm{N} \cdot \mathrm{m}\) into the torque formula. This gives us \( 5.8 = 1200 \cdot I \cdot 1.1 \times 10^{-2} \cdot 0.20 \).
04

Solve for the Current I

Re-arrange the equation to solve for the current \( I \): \[ I = \frac{5.8}{1200 \cdot 1.1 \times 10^{-2} \cdot 0.20} \].Calculate the value to find \( I \).
05

Calculation

Perform the calculation: \[ I = \frac{5.8}{1200 \cdot 0.0022} = \frac{5.8}{2.64} \approx 2.20 \text{ A} \]. Thus, the current in the coil is approximately \(2.20\) Amperes.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Magnetic Field
A magnetic field is an invisible force that is created by electric currents and magnetic materials. It is a vector field, meaning it has both a direction and a magnitude. The magnetic field plays a crucial role in the working of a DC motor. In this context, it influences the force exerted on a current-carrying coil, which in turn affects the torque produced by the motor.

When a coil carrying current is placed in a magnetic field, it experiences a force due to the magnetic field. This is because of the interaction between the magnetic field and the electric current, which causes the coil to rotate. The strength of the magnetic field is measured in Teslas (T). This force is maximized when the angle between the magnetic field lines and the normal to the coil is 90 degrees.

In our problem, a magnetic field of 0.20 T is used to help calculate the torque and ultimately find the current in the motor. Understanding magnetic fields helps predict how forces will act in different configurations, which is essential for designing efficient motors.
Torque Calculation
Torque is a measure of the rotational force on an object like a coil. It tells us how much force is needed to rotate an object around an axis. In a DC motor, the torque is generated when the current-carrying coil interacts with the magnetic field. The formula for calculating the torque (\(\tau\)) is given by:
\[t\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \]
- **\(n\)**: the number of turns in the coil (1200 turns in our problem)- **\(I\)**: the current through the coil (the value we need to find)- **\(A\)**: the area of a single turn of the coil- **\(B\)**: the strength of the magnetic field- **\(\theta\)**: the angle between the magnetic field and the area vector of the coil, which is 90 degrees when torque is maximum

For maximum torque, \(\sin(\theta)\) is equal to 1 because \(\theta = 90^{\circ}\). This simplifies our calculations. We substitute the known values into the formula and solve for \(I\). This allows us to determine how much current is required to produce a given amount of rotational force.
Electric Current
Electric current is the flow of electric charge and is measured in Amperes (A). In DC motors, electric current flows through the coils, creating a magnetic field that interacts with an external magnetic field, producing torque and causing motion.

In the given exercise, finding the electric current is crucial to understanding how the motor produces the desired torque. By using the calculated formula derived from torque calculation, \(I = \frac{\tau}{n \cdot A \cdot B}\), students can determine the current that ensures the motor operates optimally. This involves rearranging the formula to solve for the current and involves substituting all the known quantities related to torque and the magnetic field.

Knowing the amount of electric current needed helps in practical applications, like setting the right voltage or checking energy consumption in real-life motor systems, making it essential for both academic understanding and practical implementations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A charge is moving perpendicular to a magnetic field and experiences a force whose magnitude is \(2.7 \times 10^{-3} \mathrm{N}\) . If this same charge were to move at the same speed and the angle between its velocity and the same magnetic field were \(38^{\circ},\) what would be the magnitude of the magnetic force magnitude of the magnetic force that the charge would experience?

The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of 0.050 T. The \(0.20-\mathrm{kg}\) aluminum rod (length \(=1.6 \mathrm{m} )\) slides without friction down the rails at a constant velocity. How much current flows through the rod?

ssm In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5}\) T. An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{m} / \mathrm{s}\) . What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

The drawing shows two long, straight wires that are suspended from a ceiling. The mass per unit length of each wire is 0.050 \(\mathrm{kg} / \mathrm{m}\) . Each of the four strings suspending the wires has a length of 1.2 \(\mathrm{m}\) . When the wires carry identical currents in opposite directions, the angle between the strings holding the two wires is \(15^{\circ} .\) What is the current in each wire?

Two parallel rods are each 0.50 m in length. They are attached at their centers to either end of a spring (spring constant \(=150 \mathrm{N} / \mathrm{m} )\) that is initially neither stretched nor compressed. When 950 \(\mathrm{A}\) of current is in each rod in the same direction, the spring is observed to be compressed by 2.0 \(\mathrm{cm}\) . Treat the rods as long, straight wires and find the separation between them when the current is present.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.