Problem 34 - Physics

Question

Consult Interactive Solution \(21.3421 .3\) at to explore a model for solving this problem. The drawing shows a thin, uniform rod, which has a length of \(0.45 \mathrm{~m}\) and a mass of \(0.094 \mathrm{~kg}\). This rod lies in the plane of the paper and is attached to the floor by a hinge at point \(P\). A uniform magnetic field of \(0.36 \mathrm{~T}\) is directed perpendicularly into the plane of the paper. There is a current \(I=4.1 \mathrm{~A}\) in the rod, which does not rotate clockwise or counterclockwise. Find the angle \(\theta\). (Hint: The magnetic force may be taken to act at the center of gravity.)

Step-by-Step Solution

Answer

\(\theta = \tan^{-1}\left(\frac{ILB}{mg}\right)\).

Step 1: Understand the Problem

The problem involves a rod with a current inside a magnetic field. We need to find the angle \(\theta\) at which the magnetic torque balances the gravitational torque, preventing the rod from rotating.

Step 2: Identify the Forces

Identify the forces at play: the gravitational force acting at the center of gravity of the rod, and the magnetic force, which also acts at the center of gravity due to the uniform current and field.

Step 3: Calculate the Gravitational Torque

The gravitational force acts downwards at the center of the rod (0.225 m from the hinge). The gravitational torque \(\tau_g\) is \(\tau_g = mg \cdot \frac{L}{2} \cdot \cos(\theta)\). Here, \(m = 0.094\,\mathrm{kg}\), \(g = 9.8\,\mathrm{m/s^2}\), and \(L = 0.45\,\mathrm{m}\).

Step 4: Calculate the Magnetic Torque

The magnetic force is given by \(F_B = ILB\), where \(I = 4.1\,\mathrm{A}\), \(L = 0.45\,\mathrm{m}\), and \(B = 0.36\,\mathrm{T}\). The torque due to this magnetic force, acting at the center of the rod, is \(\tau_m = F_B \cdot \frac{L}{2} \cdot \sin(\theta)\).

Step 5: Set Torques Equal

For equilibrium, set the magnitudes of gravitational torque and magnetic torque equal: \(mg \cdot \frac{L}{2} \cdot \cos(\theta) = ILB \cdot \frac{L}{2} \cdot \sin(\theta)\).

Step 6: Simplify and Solve for θ

Simplify the equation: \(mg \cdot \cos(\theta) = ILB \cdot \sin(\theta)\). This leads to \(\tan(\theta) = \frac{ILB}{mg}\). Calculate \(\theta = \tan^{-1}\left(\frac{ILB}{mg}\right)\). Substitute \(I = 4.1\,\mathrm{A}\), \(L = 0.45\,\mathrm{m}\), \(B = 0.36\,\mathrm{T}\), and \(m = 0.094\,\mathrm{kg}\) to find \(\theta\).

Key Concepts

Gravitational Torque

Gravitational torque is the turning effect due to the force of gravity acting on an object. To understand it, imagine the force of gravity pulling an object down. This force acts at the object's center of gravity, creating a rotational effect when the object is hinged or fixed at one end.

Here's how to calculate gravitational torque (\(\tau_g\)):

• First, find the gravitational force, which is the mass of the object (\(m\)) multiplied by the gravitational acceleration (\(g = 9.8\,\mathrm{m/s^2}\)).
• The force acts at the object's center, which in this problem is half the length (\(\frac{L}{2}\)) of the rod from the pivot point.
• The formula for gravitational torque is \(\tau_g = mg \cdot \frac{L}{2} \cdot \cos(\theta)\), which considers the angle (\(\theta\)) between the rod and the vertical direction.

By understanding these components, you can predict how gravity will cause something to rotate.

Rotational Equilibrium

Rotational equilibrium occurs when an object remains stationary or rotates with a constant velocity, meaning the net torque acting on it is zero. In simple terms, it’s a balance between all rotating forces.

In the given exercise, the rod has to stay balanced without rotating at all. Achieving rotational equilibrium involves setting the gravitational torque equal to the magnetic torque. These forces must exactly counteract each other.

• Apply the formula for gravitational and magnetic torques.
• Set them equal: \( mg \cdot \frac{L}{2} \cdot \cos(\theta) = ILB \cdot \frac{L}{2} \cdot \sin(\theta) \).
• This equation simplifies to find the angle \(\theta\) through which these forces balance.

In this state, the rod won't tilt or turn because all torques are balanced, making it a perfect example of rotational equilibrium.

Magnetic Force

Magnetic force acts on a current-carrying conductor placed in a magnetic field. This can exert a rotational influence on the conductor, just like gravity does.

For the rod carrying a current in a magnetic field, the magnetic force (\(F_B\)) can be found using the formula \(F_B = ILB\):

• \(I\) is the current flowing through the rod.
• \(L\) is the length of the rod that the field acts upon.
• \(B\) is the magnetic field strength.

This force, like gravity, acts at the center of the rod and contributes to the torque, trying to rotate the rod. But, because it can exactly balance the gravitational torque by choosing the correct angle of inclination (\(\theta\)), equilibrium is achieved.

Center of Gravity

The center of gravity is a critical point where the total gravitational force is considered to act on an object. For uniform objects like a rod, this center is at the geometric midpoint.

Understanding the center of gravity helps in calculating torques effectively:

• For the rod in this exercise, the center of gravity is \(\frac{L}{2}\) from the hinge point.
• All torques (gravitational or magnetic) are calculated using this point as the reference for where the forces act.

It simplifies many calculations because it’s the average point of mass distribution. So calculations become more manageable as forces are considered to act through this single point. This makes it an essential concept for both linear and rotational dynamics.