/*! 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 A circular loop of wire with a r... [FREE SOLUTION] | 91Ó°ÊÓ

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A circular loop of wire with a radius of \(5.0 \mathrm{~cm}\) carries a current of 1.0 A. Another circular loop of wire is concentric with (that is, has a common center with) the first and has a radius of \(10 \mathrm{~cm}\). The magnetic field at the center of the loops is double what the field would be from the first one alone, but oppositely directed. What is the current in the second loop?

Short Answer

Expert verified
Current in the second loop is \(-4 \text{ A}\).

Step by step solution

01

Understand the Magnetic Field Produced by a Circular Loop

The magnetic field at the center of a circular loop of radius \( R \) carrying a current \( I \) can be calculated using the formula: \( B = \frac{\mu_0 I}{2R} \). We need this to find both the magnetic field contributed by each loop.
02

Calculate the Magnetic Field Due to the First Loop

For the first loop with radius \( R_1 = 5 \text{ cm} = 0.05 \text{ m} \) and current \( I_1 = 1.0 \text{ A} \), the magnetic field at the center is: \( B_1 = \frac{\mu_0 \times 1.0}{2 \times 0.05} = \frac{\mu_0}{0.1} \).
03

Analyze the Condition Given in the Problem

The magnetic field from the second loop adds to the first such that the resultant field is double but oppositely directed compared to \( B_1 \). So, \( B_2 = -2B_1 \).
04

Set Up the Equation for Magnetic Field of Second Loop

Using the formula \( B_2 = \frac{\mu_0 I_2}{2 \cdot R_2} \) for the second loop with radius \( R_2 = 10 \text{ cm} = 0.1 \text{ m} \), set \( \frac{\mu_0 I_2}{0.2} = -\frac{2\mu_0}{0.1} \).
05

Solve for the Current in the Second Loop

From the equation \( \frac{\mu_0 I_2}{0.2} = -\frac{2\mu_0}{0.1} \), solve for \( I_2 \):\[ I_2 = -2 \times \frac{0.2}{0.1} = -4 \].
06

Interpretation of Result

The negative sign indicates that the current in the second loop flows in the opposite direction to that in the first loop.

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Key Concepts

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

Circular Loop
A circular loop is a closed wired circle which can carry an electric current. This simple structure can generate a magnetic field, making it a fundamental component in studying electromagnetism. By examining circular loops, we can understand how current and magnetic fields interrelate.
A circular loop behaves like a miniature magnet when a current flows through it. The direction of the current will determine the direction of the magnetic field. In our example, we imagine two circular loops sharing the same center, creating fascinating interactions between their magnetic fields.
Knowing how to calculate and understand these magnetic fields gives us insight into how many electrical devices operate, such as motors and inductors.
Current
Current is the flow of electric charge, typically carried by moving electrons in a wire. For a circular loop, this flow of electricity creates a magnetic field around the loop. The direction and strength of this field depend on the amount of current flowing through the wire.
In our problem, we have two loops each with their own current. The first loop carries a current of 1 Ampere (A). The second loop's current is of our interest, as it affects the total magnetic field at the center.
Determining whether the current flows in the same or opposite direction between loops is crucial. It influences not only the field's magnitude but also its direction, as seen in our exercise where the second loop's current was opposite in direction.
Magnetic Field Formula
The magnetic field produced by a circular loop is quantified by the formula: \[ B = \frac{\mu_0 I}{2R} \]where \( B \) is the magnetic field, \( I \) is the current, \( R \) is the radius of the loop, and \( \mu_0 \) is the permeability of free space, a constant that characterizes the magnetic influence of the current.
This formula allows us to determine how changes in current or structure, such as the radius, affect the generated magnetic field. For instance, increasing the current will increase the magnetic field linearly, while increasing the radius will decrease the strength of the field.
Understanding this formula is essential for solving problems related to magnetic fields generated by current-carrying loops.
Electricity and Magnetism
Electricity and magnetism are interlinked physical phenomena. A change in electric fields can induce magnetic fields and vice versa. This connection forms the cornerstone of electromagnetism, influencing technologies such as transformers, electric generators, and MRI machines.
In the context of our circular loops, electricity flows through wires, generating magnetic fields around them. This is a simple demonstration of the principles of electromagnetism at play.
By calculating the resulting magnetic direction and magnitude when two such magnetic fields interact, as in our loops, we gain deeper insight into these fundamental principles of physics, highlighting how intricate and precise these interactions can be.

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Most popular questions from this chapter

1 lies on the \(x\) -axis and its north end is at \(x=+1.0 \mathrm{~cm},\) while its s… # Two identical bar magnets of negligible width are located in the \(x\) -y plane. Magnet #1 lies on the \(x\) -axis and its north end is at \(x=+1.0 \mathrm{~cm},\) while its south end is at \(x=+5.0 \mathrm{~cm}\). Magnet #2 lies on the \(y\) -axis and its north end is at \(y=+1.0 \mathrm{~cm},\) while its south end is at \(y=+5.0 \mathrm{~cm} .\) (a) In what direction would a compass point if it were located at the origin? (b) Repeat part (a) for the situation where magnet #1 is reversed in polarity. [Hint: Make a sketch of the two magnets and their individual fields at the origin.

A positive charge moves horizontally to the right across this page and enters a magnetic field directed vertically downward in the plane of the page. (a) What is the direction of the magnetic force on the charge: (1) into the page, (2) out of the page, (3) downward in the plane of the page, or (4) upward in the plane of the page? Explain. (b) If the charge is \(0.25 \mathrm{C}\), its speed is \(2.0 \times 10^{2} \mathrm{~m} / \mathrm{s},\) and it is acted on by a force of \(20 \mathrm{~N},\) what is the magnetic field strength?

What is the (a) "current" due to the electron orbiting in a circular path about the proton in a hydrogen atom? (b) What magnetic field strength does this "electron current" create at the proton location? (c) If the electron is orbiting clockwise, as viewed from above its orbital plane, what is the direction of this field? Take the orbital radius to be \(0.0529 \mathrm{nm}\). [Hint: Find the electron's period by considering the centripetal force.]

A straight, horizontal segment of wire carries a current in the \(+x\) -direction in a magnetic field that is directed in the \(-z\) -direction. (a) Is the magnetic force on the wire directed in the \((1)-x-,(2)+z-,(3)+y-,\) or (4) \(-y\) -direction? Explain. (b) If the wire is \(1.0 \mathrm{~m}\) long and carries a current of \(5.0 \mathrm{~A}\) and the magnitude of the magnetic field is \(0.30 \mathrm{~T}\), what is the magnitude of the force on the wire?

A charged particle travels undeflected through perpendicular electric and magnetic fields whose magnitudes are \(3000 \mathrm{~N} / \mathrm{C}\) and \(30 \mathrm{mT}\), respectively. Find the speed of the particle if it is (a) a proton and (b) an alpha particle. (An alpha particle is a helium nucleus-a positive ion with a double positive charge of \(+2 e .\) )

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