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A 30 -turn square coil with a mass of \(0.250 \mathrm{~kg}\) and a side length of \(0.200 \mathrm{~m}\) is hinged along a horizontal side and carries a 5.00 -A current. It is placed in a magnetic field pointing vertically downward and having a magnitude of \(0.00500 \mathrm{~T}\). Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
Answer: The angle is approximately \(18.5^{\circ}\).

Step by step solution

01

1. Determine the forces on the coil

First, we need to identify the forces acting on the coil. There are two main forces to consider: the gravitational force and the magnetic force. The gravitational force acts vertically downward on the center of mass of the coil, and its magnitude is equal to the product of the mass of the coil and the acceleration due to gravity (mg). The magnetic force acts on each side of the coil, but only the forces acting on the top and bottom sides create a torque about the hinge. The magnitude of the magnetic force on each side can be calculated using the formula: \(F_B = BIL\), where B is the magnetic field, I is the current, and L is the length of the side.
02

2. Determine the torque on the coil

The torque \(\tau\) is the turning force that causes the coil to rotate about the hinge. The torque due to the gravitational force and the magnetic force should be equal for the coil to be in equilibrium. The torque due to the gravitational force can be calculated as follows: \(\tau_g = Mg\frac{L}{2}\sin(\theta)\) where \(\theta\) is the angle the plane of the coil makes with the vertical, M is the mass of the coil, g is the acceleration due to gravity, and L/2 is the distance from the center of mass to the hinge. The torque due to the magnetic force can be calculated as: \(\tau_B = BINA\cos(\theta)\) where A is the area of the coil, and N is the number of turns.
03

3. Set up the equilibrium condition and solve for the angle

For the coil to be in equilibrium, the torques due to the gravitational force and the magnetic force must be equal: \(\tau_g = \tau_B\) Substitute the expressions for \(\tau_g\) and \(\tau_B\) and solve for the angle \(\theta\): \(Mg\frac{L}{2}\sin(\theta) = BINAL\cos(\theta)\) Divide both sides by \(\frac{ML}{2}\): \(g\sin(\theta)=2BINA\) Now, divide both sides by \(g\): \(\sin(\theta)=\frac{2BINA}{g}\) Finally, take the inverse sine of both sides to find the angle \(\theta\): \(\theta = \arcsin\left(\frac{2BINA}{g}\right)\) Substitute the given values for B=0.00500 T, I=5.00 A, N=30, A=\((0.200\,\mathrm{m})^2\), and \(g=9.81\,\mathrm{m/s^2}\): \(\theta = \arcsin\left(\frac{2(0.00500\,\mathrm{T})(5.00\,\mathrm{A})(30)(0.200\,\mathrm{m})^2}{9.81\,\mathrm{m/s^2}}\right)\) Now, calculate the value of \(\theta\): \(\theta \approx 18.5^{\circ}\) Hence, the angle that the plane of the coil makes with the vertical when the coil is in equilibrium is approximately \(18.5^{\circ}\).

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Key Concepts

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

Equilibrium of Forces
In physics, equilibrium of forces occurs when all the forces acting on an object are balanced, resulting in no net force and thus no acceleration. This concept is paramount when analyzing an object in a state of rest or moving with constant velocity. In the context of the coil problem, equilibrium is achieved when the torques acting on the coil are balanced.

To ensure equilibrium, the gravitational torque, which tries to make the coil rotate due to its weight, is counterbalanced by the magnetic torque that opposes this rotation. The condition for equilibrium, mathematically speaking, is when the sum of the clockwise torques equals the sum of the counterclockwise torques.
  • For an object in mechanical equilibrium: \[ \sum F = 0 \]
  • For torques: \[ \sum \tau = 0 \]
Understanding the equilibrium of forces helps physicists solve problems involving objects subjected to multiple forces, ensuring they remain either stationary or moving at constant velocity.
Magnetic Field
A magnetic field is a vector field surrounding magnets and electric currents, crucial in determining the force exerted on a moving charge or current-carrying wire. In this problem, the magnetic field exerts a force on the coil's current-carrying wires, affecting its orientation.

The magnetic field's strength is denoted by the symbol \(B\) and measured in Tesla (T). When a current-carrying wire is placed in a magnetic field, it experiences a force given by the formula:
\[ F_B = BIL \]
where \(B\) is the magnetic field strength, \(I\) is the current, and \(L\) is the length of the wire within the field.
  • This force is responsible for the magnetic torques that affect the coil's rotation.
  • Understanding the interaction between the magnetic field and the current is key in predicting the movement and equilibrium of the coil.
Magnetic fields play a vital role in various applications, from electric motors and transformers to navigation and data storage.
Gravitational Force
Gravitational force is the attraction between any two objects with mass. It is directed towards the center of the mass creating the gravitational pull. For the coil in the problem, this force acts downward towards the Earth's center.

The gravitational force on the coil can be calculated by the formula:
\[ F_g = mg \]
where \(m\) is the mass and \(g\) is the acceleration due to gravity, typically \(9.81 \, \mathrm{m/s^2}\). This force not only pulls the coil down but also creates a torque that attempts to alter its orientation around the pivot.
  • The gravitational torque is pivotal in understanding the coil's equilibrium as it contributes to the total torque that must be balanced by the magnetic torque.
  • By analyzing gravitational effects, one can predict and calculate the motion behaviors and steady-state positions of various objects in gravitational fields.
This concept is fundamental in fields such as physics and engineering, where weight and force analysis is critical.
Physics Problem Solving
Physics problem solving often involves identifying known and unknown variables, applying relevant physical laws, and interpreting mathematical results in real-world contexts. The coil problem, for instance, demonstrates these problem-solving techniques efficiently.

Here are the steps typically followed:
  • Identify the forces acting on the object (in this case, magnetic and gravitational forces).
  • Set up equations representing the system—torque and force equations in equilibrium problems.
  • Solve for the variable of interest, employing algebraic or calculus-based methods as needed.
  • Check the solution for physical plausibility and interpret it in terms of the problem posed.
In this exercise, calculating the angle involves solving a trigonometric equation derived from equating gravitational and magnetic torques. This approach highlights the importance of breaking down complex systems and using structured methods to achieve clarity, accuracy, and meaningful results in physics.

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Most popular questions from this chapter

A straight wire carrying a current of 3.41 A is placed at an angle of \(10.0^{\circ}\) to the horizontal between the pole tips of a magnet producing a field of \(0.220 \mathrm{~T}\) upward. The poles tips each have a \(10.0 \mathrm{~cm}\) diameter. The magnetic force causes the wire to move out of the space between the poles. What is the magnitude of that force?

A rectangular coil with 20 windings carries a current of 2.00 mA flowing in the counterclockwise direction. It has two sides that are parallel to the \(y\) -axis and have length \(8.00 \mathrm{~cm}\) and two sides that are parallel to the \(x\) -axis and have length \(6.00 \mathrm{~cm} .\) A uniform magnetic field of \(50.0 \mu \mathrm{T}\) acts in the positive \(x\) -direction. What torque must be applied to the loop to hold it steady?

A current-carrying wire is positioned within a large, uniform magnetic field, \(\vec{B}\). However, the wire experiences no force. Explain how this might be possible.

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.60 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.00 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B},\) acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0}\), of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\). d) The gold ions (represented by the black circles) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=\) \(3.27 \cdot 10^{-25}\) kg. Calculate the mass of a molybdenum ion.

The Earth is showered with particles from space known as muons. They have a charge identical to that of an electron but are many times heavier \(\left(m=1.88 \cdot 10^{-28} \mathrm{~kg}\right)\) Suppose a strong magnetic field is established in a lab \((B=0.50 \mathrm{~T})\) and a muon enters this field with a velocity of \(3.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) at a right angle to the field. What will be the radius of the resulting orbit of the muon?

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