Problem 77 - University Physics with Modern Physics

Question

A circular loop of wire with area \(A\) lies in the \(xy\)-plane. As viewed along the \(z\)-axis looking in the -\(z\)-direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\overrightarrow{B}\) is given by \(\vec{\tau}\) = D(4\(\hat{\imath}\) - 3\(\hat{\jmath}\)), where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U = -\vec{\mu}\) \(\cdot\) \(\overrightarrow{B}\) is negative. The magnitude of the magnetic field is \(B_0 = 13D/IA\). (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_x\), \(B_y\), and \(B_z\) of \(\overrightarrow{B}\).

Step-by-Step Solution

Answer

\( \vec{\mu} = -IA \hat{k} \); \( B_x = \frac{3D}{IA} \), \( B_y = \frac{4D}{IA} \), \( B_z = \frac{10D}{IA} \).

Step 1: Understanding the Magnetic Moment

The magnetic moment, \( \vec{\mu} \), of a current loop is given by \( \vec{\mu} = I \cdot \vec{A} \), where \( \vec{A} \) is the area vector perpendicular to the plane of the loop. Since the current circles clockwise and we're looking from the "-z" direction, the right-hand rule gives \( \vec{\mu} \) the negative \( z \) direction, i.e., \( \vec{\mu} = -IA \hat{k} \).

Step 2: Analyzing the Torque Equation

The torque \( \vec{\tau} \) on the loop is related to \( \vec{\mu} \) and \( \overrightarrow{B} \) by \( \vec{\tau} = \vec{\mu} \times \overrightarrow{B} \). Given \( \vec{\tau} = D(4\hat{\imath} - 3\hat{\jmath}) \), we interpret this as solving for \( \overrightarrow{B} \) using the cross product of \( \vec{\mu} = -IA \hat{k} \) and an unknown \( \overrightarrow{B} = B_x \hat{\imath} + B_y \hat{\jmath} + B_z \hat{k} \).

Step 3: Calculating the Cross Product

Calculate \( \vec{\tau} = \vec{\mu} \times \overrightarrow{B} \). The cross product results in: \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 0 & -IA \ B_x & B_y & B_z \end{vmatrix} = IA (B_y \hat{i} - B_x \hat{j}) \] This must equal \( D(4\hat{\imath} - 3\hat{\jmath}) \).

Step 4: Solving for the Magnetic Field Components

Equating the components gives us two equations: \[ IA B_y = 4D \] \[ -IA B_x = -3D \] Solving gives \( B_y = \frac{4D}{IA} \) and \( B_x = \frac{3D}{IA} \).

Step 5: Using the Magnitude of Magnetic Field

Given the magnitude \( B_0 = \frac{13D}{IA} \), use the equation: \[ B_0^2 = B_x^2 + B_y^2 + B_z^2 \] Plugging in known values: \[ \left(\frac{13D}{IA}\right)^2 = \left(\frac{3D}{IA}\right)^2 + \left(\frac{4D}{IA}\right)^2 + B_z^2 \] Solving for \( B_z \), \( B_z = \pm \frac{10D}{IA} \).

Step 6: Determining Magnetic Potential Energy Condition

Since magnetic potential energy \( U = -\vec{\mu} \cdot \overrightarrow{B} \) is negative, \( \vec{\mu} \) and \( \overrightarrow{B} \) must be in opposite directions. Since \( \vec{\mu} = -IA \hat{k} \), \( B_z \) must be positive, confirming \( B_z = \frac{10D}{IA} \).

Step 7: Result

The magnetic moment vector \( \vec{\mu} = -IA \hat{k} \). The components of the magnetic field are \( B_x = \frac{3D}{IA} \), \( B_y = \frac{4D}{IA} \), and \( B_z = \frac{10D}{IA} \).

Key Concepts

Torque on a Current Loop

Torque in physics is a measure of how much a force acting on an object causes that object to rotate. In the context of a current loop, the torque is determined by the interaction between the magnetic moment of the loop and an external magnetic field. The magnetic moment \( \vec{\mu} \) is essentially a vector that represents the strength and orientation of the loop's magnetic field. A current loop generates its own magnetic field, similar to how a magnet does, and this magnetic moment is given by \( \vec{\mu} = I \cdot \vec{A} \), where \( I \) is the current and \( \vec{A} \) is the area vector that is perpendicular to the loop's plane.
To find the torque \( \vec{\tau} \) on a current loop, we use the equation \( \vec{\tau} = \vec{\mu} \times \overrightarrow{B} \), with \( \vec{\mu} \) being the magnetic moment vector and \( \overrightarrow{B} \) the external magnetic field. The cross product here gives a vector that points in the direction that would cause the loop to rotate. For the given problem, the torque's direction and magnitude are described by \( \vec{\tau} = D(4\hat{\imath} - 3\hat{\jmath}) \). This torque causes the loop to try to align itself with the external magnetic field, just like a compass aligns itself to the Earth's magnetic field.

Magnetic Field Components

Understanding magnetic field components helps us analyze how magnetic fields interact with objects like current loops. In this exercise, the magnetic field \( \overrightarrow{B} \) has components along the \( x \), \( y \), and \( z \) axes, denoted as \( B_x \), \( B_y \), and \( B_z \) respectively.

• The equation \( \overrightarrow{B} = B_x \hat{\imath} + B_y \hat{\jmath} + B_z \hat{k} \) helps in understanding the net magnetic influence in each direction.
• For this specific problem, the components were determined by equating the calculated cross product that resulted in \( IA (B_y \hat{i} - B_x \hat{j}) \) with the given torque.

Given that the current loop was subjected to a torque \( \vec{\tau} = D(4\hat{\imath} - 3\hat{\jmath}) \), we equate to find the field components:

• \( B_y = \frac{4D}{IA} \)
• \( B_x = \frac{3D}{IA} \)

So the task involves solving these components by comparing with the provided expressions and using algebra to derive each component in relation to the constant \( D \), the current \( I \), and the area \( A \) of the loop.

Magnetic Potential Energy

Magnetic potential energy is an essential concept when dealing with magnetic forces and loops. It reflects the potential energy stored in the system due to the position and orientation of the magnetic moment of the loop relative to the magnetic field. This energy is described by the formula \( U = -\vec{\mu} \cdot \overrightarrow{B} \).
The minus sign signifies that the system minimizes energy when the magnetic moment \( \vec{\mu} \) aligns with the magnetic field \( \overrightarrow{B} \). In simpler terms, this alignment is stable and requires the least amount of energy, just like a stable object requires less energy to remain stationary.
In this exercise, it was given that the magnetic potential energy \( U \) is negative, indicating the loop's tendency to align opposite to its magnetic moment direction. We find the component \( B_z \) by considering the energy condition. Since \( \vec{\mu} = -IA \hat{k} \), having \( \overrightarrow{B} \) partially aligned in the positive \( z \) direction indicates \( B_z > 0 \). Thus we solved for \( B_z \) as \( B_z = \frac{10D}{IA} \), confirming that this value keeps the energy condition satisfied, minimizing the potential energy of the system.