Problem 63 - Fundamentals of Physics

Question

A circular loop of wire having a radius of \(8.0 \mathrm{~cm}\) carries a current of 0.20 A. A vector of unit length and parallel to the dipole moment \(\vec{\mu}\) of the loop is given by \(0.60 \hat{i}-0.80 \hat{j}\). (This unit vector gives the orientation of the magnetic dipole moment vector.) If the loop is located in a uniform magnetic field given by \(\vec{B}=\) \((0.25 \mathrm{~T}) \hat{\mathrm{i}}+(0.30 \mathrm{~T}) \hat{\mathrm{k}},\) find \((\mathrm{a})\) the torque on the loop (in unit- vector notation) and (b) the orientation energy of the loop.

Step-by-Step Solution

Answer

(a) Torque: \(-0.00072 \hat{j} - 0.0008 \hat{k}\, \text{Nm}\) (b) Orientation Energy: \(-0.0006\, \text{J}\).

Step 1: Calculate the Magnetic Dipole Moment

The magnetic dipole moment (\(\vec{\mu}\)) of a loop carrying current is given by the formula \(\vec{\mu} = I \cdot A \cdot \hat{n}\), where \(I\) is the current, \(A\) is the area of the loop, and \(\hat{n}\) is the unit vector parallel to \(\vec{\mu}\). The area \(A\) of a circular loop is \(\pi r^2\). Here, \(r = 8.0 \, \text{cm} = 0.08 \, \text{m}\) and \(I = 0.20 \, \text{A}\). Calculate the area: \[ A = \pi \times (0.08)^2 = 0.0201 \, \text{m}^2 \]Substitute the values to find \(\mu\):\[ \vec{\mu} = 0.20 \times 0.0201 \times (0.60 \hat{i} - 0.80 \hat{j}) \]\[ \vec{\mu} = 0.0024 \hat{i} - 0.0032 \hat{j} \, \text{A} \cdot \text{m}^2 \].

Step 2: Calculate Torque on the Loop

The torque \(\vec{\tau}\) experienced by a magnetic dipole in a magnetic field \(\vec{B}\) is given by the cross product: \[ \vec{\tau} = \vec{\mu} \times \vec{B} \]Substitute \(\vec{\mu} = 0.0024 \hat{i} - 0.0032 \hat{j}\) and \(\vec{B} = 0.25 \hat{i} + 0.30 \hat{k}\):\[ \vec{\tau} = (0.0024 \hat{i} - 0.0032 \hat{j}) \times (0.25 \hat{i} + 0.30 \hat{k}) \]Calculating the cross product:- \(\hat{i} \times \hat{i} = 0\)- \(\hat{i} \times \hat{k} = -\hat{j}\)- \(\hat{j} \times \hat{i} = \hat{k}\)- \(\hat{j} \times \hat{k} = \hat{i}\)Substitute and solve:\[ \vec{\tau} = 0.0024 \hat{i} \times 0.30 \hat{k} - 0.0032 \hat{j} \times 0.25 \hat{i} \]\[ \vec{\tau} = -0.00072 \hat{j} - 0.0008 \hat{k} \]Thus, \(\vec{\tau} = -0.00072 \hat{j} - 0.0008 \hat{k} \, \text{N}\cdot\text{m} \).

Step 3: Calculate Orientation Energy

The potential energy \(U\) associated with the orientation of a magnetic dipole in a magnetic field is given by:\[ U = -\vec{\mu} \cdot \vec{B} \]Substitute \(\vec{\mu} = 0.0024 \hat{i} - 0.0032 \hat{j}\) and \(\vec{B} = 0.25 \hat{i} + 0.30 \hat{k}\):\[ U = -(0.0024 \times 0.25 + 0.0032 \times 0) \]\[ U = -(0.0006 + 0) \]\[ U = -0.0006 \text{J} \].

Key Concepts

Torque

Torque plays a key role in the study of physics, especially when examining forces acting on a magnetic dipole within a magnetic field. You can think of torque as a force that causes an object to rotate. For a magnetic dipole moment, given by \(\vec{\mu} \), the torque \(\vec{\tau}\) is calculated using the cross product: \[ \vec{\tau} = \vec{\mu} \times \vec{B} \] where \(\vec{B}\) is the magnetic field. This torque acts to align the dipole moment with the magnetic field. In simple terms, the cross product measures how far from parallel the vectors are. The more perpendicular they are, the greater the torque.- In the equation, \(\vec{\tau}\) is measured in newton-meters (N·m).- Remember, when the angle between \(\vec{\mu}\) and \(\vec{B}\) is 90 degrees, the torque is at its maximum.To solve problems involving torque:

• First, identify \(\vec{\mu}\) and \(\vec{B}\).
• Perform the cross product calculation using unit vectors.
• Apply the right-hand rule to determine the direction of the torque vector.

By understanding these steps, you can effectively determine the rotational force applied to the magnetic dipole within a field.

Orientation Energy

Orientation energy, often referred to as potential energy in this context, is the energy associated with the orientation of a magnetic dipole in a magnetic field. It can be calculated with the formula:\[ U = -\vec{\mu} \cdot \vec{B} \] where \(\vec{\mu}\) is the magnetic dipole moment and \(\vec{B}\) is the magnetic field. The dot product in this formula determines how much \(\vec{\mu}\) aligns with \(\vec{B}\). - If the dipole is aligned with the field, the potential energy is at its lowest.- Conversely, if it is anti-aligned, the energy is highest.The negative sign indicates that the energy is minimized when potential energy decreases as the dipole aligns with the field. This principle is similar to a compass needle that naturally aligns with the Earth's magnetic field, moving toward a state of lower energy.To determine orientation energy:

• Compute the dot product between \(\vec{\mu}\) and \(\vec{B}\).
• Consider the angle between the vectors, which directly affects the potential energy.

Understanding orientation energy helps in recognizing how energy is deployed in magnetic physics and why dipoles behave the way they do in a magnetic field.

Cross Product

The cross product is a mathematical operation used in vector mechanics, vital for calculating forces like torque. It's distinct from the dot product and results in a vector perpendicular to the plane of the two input vectors. In our context, it's used to calculate torque from a magnetic dipole moment and a magnetic field vector:\[ \vec{\tau} = \vec{\mu} \times \vec{B} \] Key points about the cross product:- **Magnitude**: Given by \( |\vec{\mu}| |\vec{B}| \sin(\theta) \), where \(\theta\) is the angle between the vectors. The magnitude helps us understand the "strength" of the resulting vector.- **Direction**: Determined by the right-hand rule. Point your thumb in the direction of the resultant vector while curling your fingers from \(\vec{\mu}\) to \(\vec{B}\).Some useful tips:

• Identify vectors component by component.
• Apply unit vector relationships: \(\hat{i} \times \hat{i} = 0\), \(\hat{i} \times \hat{j} = \hat{k}\), and consider cyclical permutations.
• Ensure calculations are done precisely to maintain the directionality and magnitude of the result.

By understanding the cross product, you gain insight into spatial relationships and interactions between vectors, a critical aspect of vector physics and essential in electromagnetism.