Question

A circular wire loop has a radius \(R=0.12 \mathrm{~m}\) and carries a current \(i=0.10 \mathrm{~A} .\) The loop is placed in the \(x y\) -plane in a uniform magnetic field given by \(\vec{B}=-1.5 \hat{z} \mathrm{~T},\) as shown in the figure. Determine the direction and the magnitude of the loop's magnetic moment and calculate the potential energy of the loop in the position shown. If the wire loop can move freely, how will it orient itself to minimize its potential energy, and what is the value of the lowest potential energy?

Short answer

Answer: The direction of the magnetic moment is along the \(\hat{z}\) axis, and its magnitude is \(0.004524 \mathrm{~Am^2}\). The lowest potential energy is \(0.006786 \mathrm{~J}\).

Step-by-step solution

Calculate the Magnetic Moment of the Loop

The magnetic moment \(\vec{m}\) of the circular wire loop is given by \(\vec{m} = i A \hat{n}\), where \(i\) is the current, \(A = \pi R^2\) is the area enclosed by the loop, and \(\hat{n}\) is a unit vector normal to the loop plane. In this case, the loop lies in the \(xy\)-plane, so \(\hat{n}=\hat{z}\). First, let's find the area: \(A = \pi R^2 = \pi (0.12^2) = 0.04524 \mathrm{~m^2}\) Next, we will find the magnetic moment: \(\vec{m} = i A \hat{n} = 0.10 \mathrm{~A} \times 0.04524 \mathrm{~m^2} \hat{z} = 0.004524 \hat{z} \mathrm{~Am^2}\) The direction of the magnetic moment is along the \(\hat{z}\) axis, and its magnitude is \(0.004524 \mathrm{~Am^2}\).

Calculate the Potential Energy

The potential energy \(U\) of the loop in the magnetic field is given by \(U = -\vec{m} \cdot \vec{B}\). We have \(\vec{m}=0.004524 \hat{z} \mathrm{~Am^2}\) and \(\vec{B}=-1.5 \hat{z} \mathrm{~T}\). Thus, \(U = -\left(0.004524 \hat{z} \mathrm{~Am^2}\right) \cdot\left(-1.5 \hat{z} \mathrm{~T}\right) = 0.006786 \mathrm{~J}\) The potential energy of the loop in its given position is \(0.006786 \mathrm{~J}\).

Determine Orientation to Minimize Potential Energy

To minimize potential energy, the torque acting on the loop due to the magnetic field should be zero. This happens when the magnetic moment \(\vec{m}\) aligns with the magnetic field vector \(\vec{B}\). Since \(\vec{B}=-1.5 \hat{z} \mathrm{~T}\), the loop will orient itself such that \(\hat{n} = -\hat{z}\). In other words, the normal vector to the loop plane must point in the opposite direction of the \(\hat{z}\) axis.

Calculate the Lowest Potential Energy

When the loop achieves the orientation that minimizes the potential energy, the magnetic moment is parallel to the magnetic field. Therefore, the lowest possible potential energy can be calculated as: \(U_{min} = -\vec{m} \cdot \vec{B} = -\left(0.004524 \mathrm{~Am^2} \times -\hat{z}\right) \cdot\left(-1.5 \hat{z} \mathrm{~T}\right) = 0.006786 \mathrm{~J}\) The lowest potential energy is \(0.006786 \mathrm{~J}\), which is the same as the potential energy in the initial position of the loop.