91Ó°ÊÓ

A thin, uniform rod Figure \(P 7.45\) with negligible mass and length \(0.200\) \(\mathrm{m}\) is attached to the floor by less hinge at point \(P\) (Fig. horizontal spring with force \(k=4.80 \mathrm{~N} / \mathrm{m}\) connects the of the rod to a vertical wall. in a uniform magnetic field \(0.340 \mathrm{~T}\) directed into the plane figure. There is current \(I=6.50 \mathrm{~A}\) in the rod, in the direction shown. (a) Calculate the torque due to the magnetic force on the rod, for an axis at \(P\). Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? Explain. (b) When the rod is in equilibrium and makes an angle of \(53.0^{\circ}\) with the floor, is the spring stretched or compressed? (c) How much energy is stored in the spring when the rod is in equilibrium?

Step by step solution

01

Understanding Torque Due to Magnetic Force

The torque \( \tau \) due to magnetic force on a current-carrying rod is given by \( \tau = B \cdot I \cdot L \cdot \sin(\theta) \), where \( B \) is the magnetic field, \( I \) is the current, \( L \) is the length of the rod, and \( \theta \) is the angle between the rod and the direction of the magnetic field. Given in this problem are \( B = 0.340 \) T, \( I = 6.50 \) A, \( L = 0.200 \) m, and \( \theta = 90^{\circ} \) because the field is into the plane. Therefore, \( \sin(\theta) = 1 \). Thus, the torque is calculated as \( \tau = 0.340 \cdot 6.50 \cdot 0.200 \cdot 1 = 0.442 \text{ N}\cdot\text{m} \). Yes, it is correct to assume the force acts at the center of gravity when the force is uniform along the length of the rod.

02

Analyzing the Equilibrium Condition

A rod in equilibrium will have the torque from the spring balancing the magnetic torque. Given the rod is at a 53.0° angle with the floor, the component of the spring force perpendicular to the rod determines its length changes, indicating if stretched or compressed. A spring stretching would exert a restoring force opposite to the magnetic torque, meaning the spring is stretched.

03

Energy Stored in the Spring

The potential energy stored in a spring is given by \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from equilibrium. When the rod is in position, it displaces the spring by the force exerted against the magnetic torque, equated by \( k \cdot x = \tau \), where \( \tau = 0.442 \text{ N}\cdot\text{m} \) from Step 1. Thus, \( x = \frac{0.442}{4.80} \). Therefore, the energy stored is \( U = \frac{1}{2} \cdot 4.80 \cdot \left(\frac{0.442}{4.80}\right)^2 \). Calculate this to find \( U ≈ 0.0203 \text{ J} \).

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Key Concepts

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

Magnetic Field

The concept of a magnetic field is vital in this context, as it refers to a region around a magnetic material or a moving electric charge within which the force of magnetism acts. In our exercise, the magnetic field is uniform with a magnitude of 0.340 Tesla, directed into the plane of the rod. This magnetic field exerts a force on the current-carrying rod and is crucial in calculating the torque due to magnetic force. Understanding how the magnetic field interacts with the rod can help you grasp why the torque calculation includes factors like the rod's length and the current's direction.

Spring Potential Energy

Spring potential energy is the energy stored in a spring when it is displaced from its equilibrium position. It's given by the formula: \( U = \frac{1}{2} k x^2 \) where \( k \) represents the spring constant, and \( x \) is the displacement. In the exercise, we determine that the spring is stretched, which means it stores potential energy. The spring is attached to the current-carrying rod, and its displacement is a result of balancing the torque caused by the magnetic force. As such, understanding this energy helps us determine the system's behavior when the rod is at equilibrium.

Equilibrium Condition

A system is in equilibrium when all forces and torques are balanced, resulting in no net movement. For the rod in question, equilibrium is achieved when the magnetic torque and the spring's restoring torque are equal and opposite. At an angle of 53 degrees with the floor, the spring counteracts the magnetic field's influence, indicating stretching as it attempts to restore the rod to its original position. Recognizing the equilibrium condition helps explain why the spring potential energy is exact at this point and ensures the rod remains stable without rotating further.

Current-Carrying Rod

In this problem, the rod conducts an electrical current of 6.50 Amperes, which interacts with the magnetic field to create a force. A current-carrying rod within a magnetic field experiences a force due to the interaction between the current and the field lines. The rod's alignment and current direction are crucial for determining torque since the magnetic force's location and direction influence the rod's rotation. This understanding underpins the critical torque calculation.

Torque Calculation

Torque due to magnetic force is crucial in this scenario as it determines the rotational effect on the rod. The torque \( \tau \) can be calculated using the formula:\[ \tau = B \cdot I \cdot L \cdot \sin(\theta) \] where \( B \) is the magnetic field, \( I \) the current, \( L \) the length of the rod, and \( \theta \) the angle relative to the magnetic field. Here, with \( \theta = 90^\circ \), the formula simplifies as \( \sin(\theta) = 1 \). The center of gravity becomes significant as the point where magnetic force acts uniformly across the rod's length. This concept is central to understanding how the rod responds rotationally to the presence of current and the magnetic field.