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Chapter 21: Problem 57

Two circular loops of wire, each containing a single turn, have the same radius of \(4.0 \mathrm{cm}\) and a common center. The planes of the loops are perpendicular. Each carries a current of \(1.7 \mathrm{A}\). What is the magnitude of the net magnetic field at the common center?

Short Answer

Expert verified
The net magnetic field at the center is approximately \(3.79 \times 10^{-5} \mathrm{T}\).

Step by step solution

01

Understand the Setup

We have two circular loops, each with a radius of 4.0 cm. The loops are perpendicular to each other at their common center, and each carries a current of 1.7 A.
02

Formula for Magnetic Field due to a Loop

The magnetic field at the center of a single loop carrying current I with radius r is given by the formula: \( B = \frac{\mu_0 I}{2r} \), where \( \mu_0 \), the permeability of free space, is \(4\pi \times 10^{-7} \mathrm{T}\cdot\mathrm{m/A} \).
03

Calculate the Magnetic Field for Each Loop

Substitute the values into the formula for each loop. For one loop, \( B_1 = \frac{(4\pi \times 10^{-7} \mathrm{T\cdot m/A})(1.7 \mathrm{A})}{2(0.04 \mathrm{m})} = 2.68 \times 10^{-5} \mathrm{T} \). Since both loops have the same current and radius, they produce the same magnitude of magnetic field.
04

Determine the Net Magnetic Field

The magnetic fields from the perpendicular loops are perpendicular to each other, so we use the Pythagorean theorem to find the net magnetic field: \( B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{(2.68 \times 10^{-5} \mathrm{T})^2 + (2.68 \times 10^{-5} \mathrm{T})^2} \).
05

Calculate the Net Magnetic Field

Calculate the numeric value: \( B_{net} = \sqrt{2(2.68 \times 10^{-5} \mathrm{T})^2} = \sqrt{2}(2.68 \times 10^{-5} \mathrm{T}) \approx 3.79 \times 10^{-5} \mathrm{T} \).

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Key Concepts

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

Circular Loops
Circular loops of wire play a crucial role in magnetic field experiments and applications. When a current passes through a loop of wire, it creates a magnetic field. The shape and orientation of the circular loops determine the nature of this magnetic field. In the exercise, two loops are considered, each having a radius of 4.0 cm. Keep in mind:
  • A circular loop is essentially a coil made from a wire.
  • The area inside the loop affects the strength of the magnetic field.
  • The orientation of the loop (horizontal, vertical, etc.) impacts the direction of the magnetic field.
In our exercise, the loops are positioned perpendicularly, meaning they intersect at 90 degrees, influencing the resultant magnetic field at their common center. This perpendicular positioning is essential because it ensures that each loop's magnetic field contributes independently to the final magnetic field.
Current
Current is the flow of electric charge and is measured in Amperes (A). It is a vital factor determining the strength of the magnetic field produced by a circular loop. In our specific problem, each loop carries a current of 1.7 A. This current flows uniformly through the loops, creating magnetic fields:
  • Current direction determines the orientation of the magnetic field via the right-hand rule.
  • The intensity of the current influences the strength of the magnetic field.
The right-hand rule is a helpful tool: point your thumb in the current's direction, and your fingers will curl in the direction of the magnetic field lines. This understanding helped calculate the magnitude of magnetic fields produced by each loop.
Perpendicular Planes
When planes are perpendicular, they meet at a right angle (90 degrees). This alignment allows the magnetic fields produced by individual currents in our loops to be considered independently. Here's why this matters:
  • Magnetic fields are vector quantities, meaning they have both magnitude and direction.
  • When objects are perpendicular, their effects can be calculated separately and combined using the Pythagorean theorem.
In the exercise, each loop creates a magnetic field perpendicular to the other. Thus, to find the net magnetic field at the center, we apply the Pythagorean theorem: \[B_{net} = \sqrt{B_1^2 + B_2^2}\] This equation allows us to combine the magnetic fields to find their resultant strength at the perpendicular intersection.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a fundamental constant in physics. It describes how well a magnetic field can penetrate a vacuum and is critical in calculating magnetic fields from currents. For free space, \( \mu_0 = 4\pi \times 10^{-7} \mathrm{T}\cdot\mathrm{m/A} \).
It plays a direct role in determining the strength of the magnetic field inside a loop:
  • Appears in the formula for magnetic fields due to currents: \( B = \frac{\mu_0 I}{2r} \) This formula is used when calculating magnetic fields produced by loops like the ones in our problem.
  • It's a constant that remains the same across all situations involving magnetic fields in a vacuum.
This parameter helps link the physical dimensions of the loop and the current with the resultant magnetic field strength, thus completing the calculation of the field in a systematic manner.

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Most popular questions from this chapter

Two insulated wires, each \(2.40 \mathrm{m}\) long, are taped together to form a two-wire unit that is \(2.40 \mathrm{m}\) long. One wire carries a current of \(7.00 \mathrm{A}\); the other carries a smaller current \(I\) in the opposite direction. The twowire unit is placed at an angle of \(65.0^{\circ}\) relative to a magnetic field whose magnitude is \(0.360 \mathrm{T}\). The magnitude of the net magnetic force experienced by the two-wire unit is \(3.13 \mathrm{N}\). What is the current \(I ?\)

Two infinitely long, straight wires are parallel and separated by a distance of one meter. They carry currents in the same direction. Wire 1 carries four times the current that wire 2 carries. On a line drawn perpendicular to both wires, locate the spot (relative to wire 1 ) where the net magnetic field is zero. Assume that wire 1 lies to the left of wire 2 and note that there are three regions to consider on this line: to the left of wire \(1,\) between wire 1 and wire \(2,\) and to the right of wire 2.

At New York City, the earth's magnetic field has a vertical component of \(5.2 \times 10^{-5} \mathrm{T}\) that points downward (perpendicular to the ground) and a horizontal component of \(1.8 \times 10^{-5} \mathrm{T}\) that points toward geographic north (parallel to the ground). What are the magnitude and direction of the magnetic force on a \(6.0-\mathrm{m}\) long, straight wire that carries a current of 28 A perpendicularly into the ground?

A piece of copper wire has a resistance per unit length of \(5.90 \times 10^{-3} \Omega / \mathrm{m} .\) The wire is wound into a thin, flat coil of many turns that has a radius of \(0.140 \mathrm{m}\). The ends of the wire are connected to a \(12.0-\mathrm{V}\) battery. Find the magnetic field strength at the center of the coil.

In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5} \mathrm{T} .\) 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.
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