/*! 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 45 A coil consists of 120 circular ... [FREE SOLUTION] | 91Ó°ÊÓ

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A coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

Short Answer

Expert verified
Solution: Using the given values and calculations, we can find the strength of the magnetic field by plugging them into the equation, \(B = \frac{Ï„}{M}\): \(B = \frac{1.2 \cdot 0.048}{120 \cdot 0.49 \cdot \pi(0.048)^2}\) Upon solving, we get: \(B \approx 0.201 \, T\) Thus, the strength of the magnetic field acting on the circular coil is approximately \(0.201 \, T\).

Step by step solution

01

Calculate the magnetic moment (M) of the coil

The magnetic moment (M) of a coil is given by the product of the current (I), the area of the loop (A), and the number of loops (n): \(M = nIA\) We are given: - n = 120 loops - I = 0.49 A - A = πr^2, radius (r) = 4.8 cm = 0.048 m We can plug these values into the formula to find M: \(M = 120 \cdot 0.49 \cdot \pi(0.048)^2\)
02

Calculate the torque (Ï„) experienced by the coil

The torque experienced by a coil is given by the force (F) applied to keep the coil from rotating multiplied by the perpendicular distance (r) from the rotation axis: \(Ï„ = Fr\) Here, we have: - F = 1.2 N - r = 0.048 m We can now calculate the torque: \(Ï„ = 1.2 \cdot 0.048\)
03

Relate the torque (Ï„) to the magnetic moment (M) and the magnetic field (B)

The torque acting on a coil is also given by the product of the magnetic moment (M) and the magnetic field (B): \(τ = MB\sin\theta\) Since the coil is oriented parallel to the \(x\)-axis, the angle θ between the magnetic field and the plane of the coil is \(90^{\circ}\). The sine of \(90^{\circ}\) is 1, so we have: \(τ = MB\) Now we have enough information to solve for the magnetic field (B).
04

Calculate the magnetic field (B)

We can write the expression for B as: \(B = \frac{Ï„}{M}\) We can substitute the values obtained in steps 1 and 2 to find the magnetic field: \(B = \frac{1.2 \cdot 0.048}{120 \cdot 0.49 \cdot \pi(0.048)^2}\) Now, solve for B to get the strength of the magnetic field.

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Key Concepts

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

Torque Calculation
Understanding torque is essential when analyzing how coils interact in a magnetic field. Torque (\(\tau\)) in relation to a coil is the measure of the force that causes the coil to rotate. It acts around the axis of rotation. Here, torque is determined by multiplying the magnitude of the force applied (\(F = 1.2 \, \text{N}\)) by the perpendicular distance from the axis to the point where the force is applied (\(r = 0.048 \, \text{m}\)).
In this scenario, the formula \(\tau = Fr\) helps us find the exact torque acting on the coil. This simple yet crucial computation allows us to understand how much force is needed to maintain a certain rotational equilibrium. When torque is applied, it either causes the coil to begin rotating or keeps it from rotating, depending on external conditions. Calculating torque gives insights into the interplay between physical forces and their ability to induce motion in conductive materials exposed to force fields.
Magnetic Field Strength
The magnetic field (\(B\)) represents a vector field around a magnetic material or a moving electric charge where the magnetic forces are observable. Here, we relate the magnetic field to both the torque on the coil and the coil’s magnetic moment.
From the equation \(\tau = MB\sin \theta\), where \(M\) is the magnetic moment and \(\theta\) is the angle, we recognize that the torque (\(\tau\)) is directly proportional to the magnetic field strength (\(B\)). Since the coil is aligned parallel to the \(x\)-axis, meaning the angle \(\theta = 90^{\circ}\), and since \(\sin 90^{\circ} = 1\), the equation simplifies to \(\tau = MB\).By rearranging and solving \(B = \frac{\tau}{M}\), we determine the magnetic field strength, which shows how the field influences the coil and affects its ability to remain stationary or rotate.
Coil Rotation
Coil rotation in a magnetic field offers a fascinating physical phenomenon. A coil, when subjected to a magnetic field, experiences a force that could lead to its rotation. This rotational motion is influenced by several factors including the torque, magnetic field strength, and orientation of the coil.
In the given problem, a force prevents the coil from rotating despite the presence of a magnetic field. This scenario illustrates a balance of forces, where torque counteracts rotational movement. It points out an essential concept of equilibrium. An essential condition for torque to achieve maximum value, subsequently affecting the coil's rotation, is that the applied force and magnetic field direction form a significant angle. Here, when aligned parallelly (at \(90^{\circ}\)), the coil experiences maximum torque. Thus, understanding coil dynamics aids not just in problem-solving but also expands conceptual knowledge of how mechanical and electromagnetic forces synchronize.
Current in a Coil
The current in a coil plays a critical role in determining its magnetic properties. It is directly linked with the magnetic moment, which describes the tendency of the coil to align with a magnetic field.
In this exercise, the coil carries a current of \(0.49 \, \text{A}\). The magnetic moment (\(M\)) of the coil can be calculated as the product of current (\(I\)), the area of the loop (\(A\)), and the number of loops (\(n\)). Utilizing the equation \(M = nIA\), provides insight into how electricity flow within the coil interacts with external forces.Current also affects the magnitude of the torque experienced by the coil, as it determines the strength of the coil's own magnetic field. This interplay allows the coil to either resist or succumb to external magnetic influences, thus affecting its rotational dynamics. Understanding current's impact is vital for conceptualizing electromagnetic machines and devices.

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Most popular questions from this chapter

An alpha particle \(\left(m=6.6 \cdot 10^{-27} \mathrm{~kg}, q=+2 e\right)\) is accelerated by a potential difference of \(2700 \mathrm{~V}\) and moves in a plane perpendicular to a constant magnetic field of magnitude \(0.340 \mathrm{~T}\), which curves the trajectory of the alpha particle. Determine the radius of curvature and the period of revolution.

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

A helium leak detector uses a mass spectrometer to detect tiny leaks in a vacuum chamber. The chamber is evacuated with a vacuum pump and then sprayed with helium gas on the outside. If there is any leak, the helium molecules pass through the leak and into the chamber, whose volume is sampled by the leak detector. In the spectrometer, helium ions are accelerated and released into a tube, where their motion is perpendicular to an applied magnetic field, \(\vec{B},\) and they follow a circular orbit of radius \(r\) and then hit a detector. Estimate the velocity required if the orbital radius of the ions is to be no more than \(5 \mathrm{~cm},\) the magnetic field is \(0.15 \mathrm{~T}\) and the mass of a helium- 4 atom is about \(6.6 \cdot 10^{-27} \mathrm{~kg}\). Assume that each ion is singly ionized (has one electron less than the neutral atom). By what factor does the required velocity change if helium- 3 atoms, which have about \(\frac{3}{4}\) as much mass as helium- 4 atoms, are used?

It would be mathematically possible, for a region with zero current density, to define a scalar magnetic potential analogous to the electrostatic potential: \(V_{B}(\vec{r})=-\int_{\vec{r}_{0}}^{\vec{r}} \vec{B} \cdot d \vec{s},\) or \(\vec{B}(\vec{r})=-\nabla V_{B}(\vec{r}) .\) However, this has not been done. Explain why not.

Initially at rest, a small copper sphere with a mass of \(3.00 \cdot 10^{-6} \mathrm{~kg}\) and a charge of \(5.00 \cdot 10^{-4} \mathrm{C}\) is accelerated through a \(7000 .-\mathrm{V}\) potential difference before entering a magnetic field of magnitude \(4.00 \mathrm{~T}\), directed perpendicular to its velocity. What is the radius of curvature of the sphere's motion in the magnetic field?

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