Problem 85 - University Physics with Modern Physics

Question

A circular loop of wire with area \(A\) lies in the \(x y\) -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 extemal magnetic field \(\vec{B}\) is given by \(\vec{\tau}=D(4 \hat{z}-3 \hat{y}),\) where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U=-\vec{\mu} \cdot \vec{B}\) is negative. The magnitude of the magnetic field is \(B_{0}=13 D / L A\) (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_{x}, B_{y}\) , and \(B_{z}\) of \(\vec{B}\) .

Step-by-Step Solution

Answer

(a) \( \vec{\mu} = \mu \hat{z}, \text{magnitude } \mu = \frac{3D}{B_x} \). (b) \(B_x = \text{\text{solve force equ potential}} \( B_y = 0, B_z = B_0\).}

Step 1: Understanding Torque and Magnetic Moment

In this exercise, we need to relate the torque \( \vec{\tau} \) experienced by a loop of wire in a magnetic field to its magnetic moment \( \vec{\mu} \) and the magnetic field \( \vec{B} \). The relationship is \( \vec{\tau} = \vec{\mu} \times \vec{B} \). We're given that \( \vec{\tau} = D(4 \hat{z} - 3 \hat{y}) \). The magnetic moment \( \vec{\mu} \) is associated with the current and the geometry of the loop.

Step 2: Determining the Magnetic Moment

Since the torque is given by the expression \( \vec{\tau} = \vec{\mu} \times \vec{B} \), we know \( \vec{\tau} \) is perpendicular to both \( \vec{\mu} \) and \( \vec{B} \). Given the loop's alignment in the \( xy \)-plane and current's clockwise direction from \(-z\), the magnetic moment \( \vec{\mu} \) is directed along \(\hat{z}\). Hence, \( \vec{\mu} = \mu \hat{z} \).

Step 3: Solve for Magnetic Moment Magnitude

Using the equation for torque, \( D(4\hat{z} - 3\hat{y}) = \vec{\mu} \times \vec{B} \), we assume \( \vec{B} = B_x \hat{x} + B_y \hat{y} + B_z \hat{z} \). The cross-product expression simplifies to \(-\mu B_y \hat{x} + \mu B_x \hat{y} + 0\, \hat{z}\) = \(D(4\hat{z} - 3\hat{y})\). Therefore, \( \mu B_x=3D \) and \( \mu B_y=0 \).

Step 4: Solving for Magnetic Field Components

The magnitude of \( \vec{B} \) is \( B_0 = \frac{13D}{LA} \). Checking for consistency with \( \mu B_x=3D \) and \( \mu B_y=0 \), we also use energy \( U = -\vec{\mu} \cdot \vec{B} < 0 \). Thus, \( \vec{\mu} = 3D/ B_x \hat{z} \) is negative if \( B_z > 0 \) and negative with \( \mu \).

Step 4: Express Magnetic Field Components

The vector \( \mu \) across \(\hat{x}, \hat{y}, \hat{z}\) terms imply cartesian bounds on \( \vec{B} \) so that \( B_x = \frac{3D}{\mu}, B_y = 0, B_z = B_0\). The \( z\)-component maintains energy condition under field affect.

Key Concepts

Torque and Magnetic Field

When dealing with a current loop in a magnetic field, one important concept is the torque produced on the loop. Torque is a twisting force that tends to cause rotation. The exact way torque works here involves both the loop's magnetic moment and the magnetic field itself. The equation that describes this relationship is \( \vec{\tau} = \vec{\mu} \times \vec{B} \). Here, \( \vec{\tau} \) represents torque, \( \vec{\mu} \) is the magnetic moment, and \( \vec{B} \) is the magnetic field. The cross product (\( \times \)) signifies that the torque is perpendicular to both the magnetic moment and the magnetic field.

• Torque depends on the orientation of the magnetic moment relative to the magnetic field. If they are aligned, torque is minimized.
• In our exercise, the torque is given as \( D(4 \hat{z} - 3 \hat{y}) \), implying specific aligns of the components.

The magnitude of torque will give you insight into how the loop tries to rotate within the field. It's a crucial part of understanding magnetic interactions.

Current Loop

Current loops are an essential part of electromagnetism and physics. A current loop is simply a loop made of wire through which electric current flows. The current loop in this exercise lies flat in the \( xy \)-plane, forming a vital part of understanding the magnetic moment.Any loop carrying a current will have a magnetic moment \( \vec{\mu} \). The direction of this magnetic moment is given by the right-hand rule. For a current loop viewed from the positive \( z \)-axis as moving clockwise, the magnetic moment extends along the negative \( z \)-axis.

• The magnetic moment is proportional to the current \( I \) and the area \( A \) of the loop: \( \vec{\mu} = I \times A \times \hat{n} \), where \( \hat{n} \) is the unit vector perpendicular to the loop.
• In this case, because the loop lies in the \( xy \)-plane, \( \vec{\mu} \) aligns with the \( z \)-axis.

The current loop is central to deriving outcomes like torque or the interactions of the loop in fields. These loops underpin a lot of electromagnetic concepts you'll encounter.

Cross Product in Magnetostatics

The cross product is a key mathematical tool used extensively in physics to handle vector quantities, especially in magnetostatics. It helps determine directions and magnitudes useful in understanding how magnetic moments and fields interact.In the expression \( \vec{\tau} = \vec{\mu} \times \vec{B} \), the cross product \( \times \) calculates the torque by determining the perpendicular vector resulting from \( \vec{\mu} \) and \( \vec{B} \).

• The direction of the resultant vector (like torque) is perpendicular to the plane formed by the two original vectors.
• The magnitude of this vector is the product of the magnitudes of the two vectors and the sine of the angle between them.

This math insight helps solve physics problems by clearly showing the resultant directions, especially when analyzing forces and fields in electromagnetism. The cross product not only provides direction but helps deduce component alignments as seen in the exercise.