91Ó°ÊÓ

A straight piece of conducting wire with mass \(M\) and length \(L\) is placed on a frictionless incline tilted at an angle \(\theta\) from the horizontal (as shown in Fig. 9.312). There is a uniform, vertical magnetic field at all points (produced by an arrangement of magnets not shown in Fig. 9.312). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to rernain at rest. a. \(\frac{\mathrm{Mg} \tan \theta}{2 L B}\) to the left b. \(\frac{\mathrm{Mg} \tan \theta}{L B}\) to the right c. \(\frac{\mathrm{Mg} \tan \theta}{L B}\) to the left d. \(\frac{3 \mathrm{Mg} \tan \theta}{2 L B}\) to the left

Step by step solution

01

Analyze the Forces Acting on the Wire

Firstly, consider the forces acting on the wire. The wire is on a frictionless incline, so the gravitational force will cause it to slide down the incline. The force due to gravity acting down the incline is given by the parallel component of the gravitational force: \[ F_{gravity} = Mg \sin \theta \] where \( M \) is the mass of the wire, and \( g \) is the acceleration due to gravity. The magnetic force due to the current in the wire will act perpendicular to the direction of the magnetic field and the current.

02

Use the Right-Hand Rule to Determine Magnetic Force Direction

Apply the right-hand rule to find the direction of the magnetic force. For a wire with current flowing in a magnetic field, the force direction is perpendicular to both the direction of the field and the current. If the magnetic field is vertical and the wire is on the incline, the force will be exerted either up or down the incline. To keep the wire stationary, the magnetic force should counteract the gravitational force, acting upward along the incline.

03

Expression for Magnetic Force

The force on a current-carrying wire in a magnetic field is given by the formula:\[ F_{magnetic} = ILB \cos \theta \] where \( I \) is the current through the wire, \( L \) is the length of the wire, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the wire and the magnetic field. In this case, since the current needs to oppose gravity, it should be such that it balances the gravitational force.

04

Equate Forces to Find the Required Current

To keep the wire stationary, the magnetic force must equal the gravitational force:\[ ILB \cos \theta = Mg \sin \theta \]Rearranging the equation for the current, \( I \):\[ I = \frac{Mg \sin \theta}{LB} \]Since \( \sin \theta \) for the force that pushes back up the incline equals \( \tan \theta \cos \theta \) (since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)), we can rewrite it as:\[ I = \frac{Mg \tan \theta}{LB} \]

05

Determine the Direction of the Current

Since the right-hand rule indicates that to produce a magnetic force up the incline, the current must flow in a specific direction, apply the right-hand grip rule or similar to determine the current's direction. With the magnetic field directed vertically, to counteract gravity, the current should flow left if the directional outcome from previous force analysis rules is considered.

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Key Concepts

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

Current in Conducting Wire

When we talk about current in a conducting wire, we are referring to the flow of electrons through the wire. This movement of charges is what creates the current. In physics, especially electromagnetism, current flow can dramatically interact with external environments like magnetic fields.

In this exercise, the wire's current is crucial for balancing forces on an incline. A voltage source is attached to the wire, enabling a controlled current to flow, which interacts with the magnetic field present. The direction and magnitude of this current are essential to create a magnetic force that can counterbalance gravity's pull on the wire due to its inclination. Understanding how this current affects the wire's motion is key to solving physics problems involving forces and electromagnetism.

Forces on Incline

When a wire rests on an incline, several forces act upon it. Here, the incline is frictionless, which alters the typical force dynamics. Gravitational force is the primary acting force, pulling the wire down the slope.

The gravitational force can be broken down into two components when dealing with an incline:

• Perpendicular to the incline: does not affect motion along the slope.
• Parallel to the incline: results in motion down the slope, calculated as \( Mg \sin \theta \) where \( \theta \) is the angle of the incline.

To keep the wire from sliding, another opposing force, namely the magnetic force due to the wire's current, must counteract this gravitational pull.

Right-Hand Rule

The right-hand rule is a powerful tool for determining the direction of the magnetic force on a current-carrying wire. When a wire carrying current is placed in a magnetic field, the interaction between the current and magnetic field produces a force.

To apply the right-hand rule, follow these steps:

• Point your thumb in the direction of the current in the wire.
• Extend your fingers in the direction of the magnetic field.
• Your palm will then face in the direction of the force exerted on the wire.

In this problem, applying the right-hand rule helps determine that the magnetic force needs to act up the incline. Hence, the current must flow in such a direction that accomplishes this opposite force.

Equilibrium of Forces

A state of equilibrium occurs when all forces acting on an object are balanced, resulting in no net force and hence no motion. In this scenario, the magnetic force due to the current must balance the gravitational force tugging the wire down the incline for the wire to remain stationary.

For equilibrium to be achieved:

• The magnetic force (\( ILB \cos \theta \)) must equal the gravitational component (\( Mg \sin \theta \)).
• Solving \( ILB \cos \theta = Mg \sin \theta \) for \( I \), gives \( I = \frac{Mg \tan \theta}{LB} \).

Such balance ensures that the magnetic force perfectly offsets the gravitational pull, keeping the wire at rest. Mastery of force equilibrium allows physicists to predict and control systems influenced by multiple forces.