/*! 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 62 \(\bullet\) Two circular concent... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet\) Two circular concentric loops of wire lie on a tabletop, one inside the other. The inner loop has a diameter of 20.0 \(\mathrm{cm}\) and carries a clockwise current of 12.0 \(\mathrm{A}\) , as viewed from above, and the outer wire has a diameter of 30.0 \(\mathrm{cm} .\) What must be the magnitude and direction (as viewed from above) of the current in the outer loop so that the net magnetic field due to this combination of loops is zero at the common center of the loops?

Short Answer

Expert verified
18.1 A, counterclockwise, in the outer loop.

Step by step solution

01

Identify Formula for Magnetic Field of Circular Loop

The magnetic field at the center of a single circular loop carrying a current is given by the formula \( B = \frac{\mu_0 I}{2R} \), where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \: \text{T}\cdot \text{m}/\text{A} \)), \( I \) is the current, and \( R \) is the radius of the loop.
02

Calculate the Magnetic Field of the Inner Loop

The radius of the inner loop is half of its diameter: \( R_{\text{inner}} = 10.0 \: \text{cm} = 0.10 \: \text{m} \). Substituting into the magnetic field formula gives: \( B_{\text{inner}} = \frac{(4\pi \times 10^{-7}) \times 12.0}{2 \times 0.10} \). Calculating this gives \( B_{\text{inner}} = 7.54 \times 10^{-5} \: \text{T} \).
03

Set the Net Magnetic Field to Zero

For the net magnetic field at the center to be zero, the magnetic field produced by the outer loop must equal in magnitude but opposite in direction to that of the inner loop. Therefore, \( B_{\text{outer}} = -B_{\text{inner}} = -7.54 \times 10^{-5} \: \text{T} \).
04

Calculate the Current in the Outer Loop

Using the magnetic field formula for the outer loop \( B = \frac{\mu_0 I_{\text{outer}}}{2R_{\text{outer}}} \), where \( R_{\text{outer}} = 15.0 \: \text{cm} = 0.15 \: \text{m} \), substitute in the values: \(-7.54 \times 10^{-5} = \frac{4\pi \times 10^{-7} I_{\text{outer}}}{2 \times 0.15}\). Solving for \( I_{\text{outer}} \) gives \( I_{\text{outer}} = -18.1 \: \text{A} \).
05

Determine the Direction of the Outer Current

Since the current in the inner loop is clockwise and creates a downward magnetic field, to cancel this out the outer loop must create an upward field at the center. The right-hand rule indicates that a current counterclockwise when viewed from above will generate an upward magnetic field at the center.

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.

circular_loops
Circular loops are loops which have a consistent circular shape. They are essential in physics, especially when discussing magnetic fields. When a current passes through a circular loop of wire, it generates a magnetic field perpendicular to the plane of the loop. Circular loops are commonly used in experiments and applications to demonstrate how electricity and magnetism interact.
In this context, we have two concentric circular loops, meaning one loop is inside the other, each lying flat on a tabletop. The key aspect for these loops is their size, expressed by their diameter. The inner loop has a diameter of 20.0 cm, while the outer is larger, at 30.0 cm.
The magnetic field generated by a circular loop depends heavily on its size and the current flowing through it. Knowing the radii helps in calculating the magnetic field strength at the center.
current_direction
The direction of the current plays a crucial role in determining the direction of the magnetic field it produces. In our example, the inner loop carries a clockwise current when seen from above. Current direction is not arbitrary; it is determined by how the electrons move through the conducting path.
Direction of current decides the loop's magnetic field direction due to the magnetic field lines rotating around the wire path. If you invert the current direction, the magnetic field direction also flips.
In the exercise, we're given that the inner loop has this clockwise current, while the calculation for the outer loop suggests that it should carry a current counterclockwise. This opposing direction is necessary to cancel out the magnetic fields at the center of the loops.
right_hand_rule
The right-hand rule is a mnemonic that helps determine the direction of the magnetic field relative to the direction of current flow. To apply it, you "grab" the loop of wire with your right hand such that your thumb points in the direction of the current flow. Your fingers then curl in the direction of the generated magnetic field lines.
Using this rule, we decide the necessary direction for current in the outer loop. For the inner loop, with clockwise current, the magnetic field at the center goes downward. To cancel this field at the center, the current in the outer loop must create an upward magnetic field.
This upward field is generated when the current in the outer loop is counterclockwise, as the right-hand rule indicates that a counterclockwise current results in an upward field at the loop's center when viewed from above.
net_magnetic_field
The net magnetic field is the sum of all individual magnetic fields present at a certain point. In our case, it's the result of adding the fields from two loops. We're tasked with adjusting these fields to make the net magnetic field at the loops' common center zero.
To achieve this, the magnetic fields produced by each loop must be equal in magnitude but opposite in direction. The inner loop creates a downward magnetic field due to its clockwise current. Thus, the task is to engineer the outer loop's field to be upward and equal in strength.
By adjusting the current magnitude and direction in the outer loop, specifically to -18.1 A in a counterclockwise direction, the net magnetic field at the center can effectively be nulled, as both fields perfectly counterbalance each other.

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

\(\bullet\) A singly charged ion of \(^{7} \mathrm{Li}\) (an isotope of lithium containing three protons and four neutrons) has a mass of \(1.16 \times\) \(10^{-26} \mathrm{kg} .\) It is accelerated through a potential difference of 220 \(\mathrm{V}\) and then enters a 0.723 T magnetic field perpendicular to the ion's path. What is the radius of the path of this ion in the magnetic field?

\(\bullet\) As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near its center. You have enough wire for 4000 circular turns, and the solenoid must be 1.40 m long and 2.00 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

\(\bullet\) A closely wound circular coil has a radius of 6.00 \(\mathrm{cm}\) and carries a current of 2.50 A. How many turns must it have if the magnetic field at its center is \(6.39 \times 10^{-4} \mathrm{T} ?\)

\(\bullet\) Atom smashers! A cyclotron particle accelerator (sometimes called an "atom smasher" in the popular press) is a device for accelerating charged particles, such as electrons and protons, to speeds close to the speed of light. The basic design is quite simple. The particle is bent in a circular path by a uniform magnetic field. An electric field is pulsed periodically to increase the speed of the particle. The charged particle (or ion) of mass \(m\) and charge \(q\) is introduced into the cyclotron so that it is moving perpendicular to a uniform magnetic field \(\vec{B}\) (a) Starting with the radius of the circular path of a charge moving in a uniform magnetic field, show that the time \(T\) for this particle to make one complete circle is \(T=\frac{2 \pi m}{|q| B}\) . (Hint: You can express the speed \(v\) in terms of \(R\) and \(T\) because the particle travels through one circumference of the circle in time \(T\) . (b) Which would take longer to complete one circle, an ion moving in a large circle or one moving in a small circle? Explain.

A particle with a charge of \(-2.50 \times 10^{-8} \mathrm{C}\) is moving with an instantaneous velocity of magnitude 40.0 \(\mathrm{km} / \mathrm{s}\) in the \(x y-\) plane at an angle of \(50^{\circ}\) counterclockwise from the \(+x\) axis. What are the magnitude and direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the (a) \(-x\) direction, and (b) \(+z\) direction?

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.