Question

Show that the differential work in moving a current element \(I d \mathbf{L}\) through a distance \(d \mathbf{l}\) in a magetic field \(\mathbf{B}\) is the negative of that done in moving the element \(I d \mathbf{l}\) through a distance \(d \mathbf{L}\) in the same field.

Short answer

Question: Show that the differential work done in moving a current element \(Id\textbf{L}\) through a distance \(d\textbf{l}\) in a magnetic field \(\textbf{B}\) is equal but opposite in sign (negative) as that done in moving the element \(Id\textbf{l}\) through a distance \(d\textbf{L}\) in the same magnetic field. Answer: Using the cross product property, we have shown that the differential work in both cases (\(\bold{W_1}\) and \(\bold{W_2}\)) are negatives of each other, with \(\bold{W_1} = -\bold{W_2}\). Therefore, the differential work done in moving a current element \(Id\textbf{L}\) through a distance \(d\textbf{l}\) in a magnetic field \(\textbf{B}\) is equal but opposite in sign (negative) as that done in moving the element \(Id\textbf{l}\) through a distance \(d\textbf{L}\) in the same magnetic field.

Step-by-step solution

Calculate the force on a current-carrying wire in a magnetic field

The force \(\textbf{F}\) on a current-carrying wire of length \(\Delta \bold{l}\) and current \(I\) in a magnetic field \(\bold{B}\) can be obtained using the formula: $$ \textbf{F} = I(\Delta \bold{l} \times \bold{B}). $$

Calculate the differential work done in moving a current element through a distance in a magnetic field

Differential work \(\bold{W}\) is the dot product of the force acting on the object and the displacement vector (\(\Delta \bold{l}\) or \(\Delta \bold{L}\)) of the object. Therefore, the differential work done in moving the current element \(I \bold{dL}\) through a distance \(\bold{dl}\) can be written as: $$ \bold{W_1} = \bold{F_1} \cdot \bold{dl} $$ where \(\bold{F_1} = I(\bold{dL} \times \bold{B})\) represents the force acting on the current element. Substitute the value of \(\bold{F_1}\) into the equation for \(\bold{W_1}\), we have: $$ \bold{W_1} = I(\bold{dL} \times \bold{B}) \cdot \bold{dl}. $$

Calculate work done in moving current element through another distance in the same magnetic field

Now, we need to find the work done in moving the current element \(I\bold{dl}\) through a distance \(\bold{dL}\). This work can be represented as: $$ \bold{W_2} = \bold{F_2} \cdot \bold{dL} $$ where \(\bold{F_2} = I(\bold{dl} \times \bold{B})\) represents the force acting on the current element. Substitute the value of \(\bold{F_2}\) into the equation for \(\bold{W_2}\), we have: $$ \bold{W_2} = I(\bold{dl} \times \bold{B}) \cdot \bold{dL}. $$

Show that the differential work in both cases are negatives of each other

Now that we have expressions for the differential work done in both cases, we need to show that \(W_1 = -W_2\). We can do this by using the properties of the dot product and cross product. $$ \bold{W_1} = I(\bold{dL} \times \bold{B}) \cdot \bold{dl} \\ \bold{W_2} = I(\bold{dl} \times \bold{B}) \cdot \bold{dL} $$ Using the cross product property \((\bold{a}\times\bold{b})\cdot\bold{c} = (\bold{b}\times\bold{c})\cdot\bold{a}\), we can rewrite the work done in both cases as: $$ \bold{W_1} = I(\bold{B} \times \bold{dl}) \cdot \bold{dL} \\ \bold{W_2} = I(\bold{dl} \times \bold{B}) \cdot \bold{dL} $$ From this, we can say that the differential work done in both cases (\(\bold{W_1}\) and \(\bold{W_2}\)) are negatives of each other: $$ \bold{W_1} = -\bold{W_2} $$ This shows that the differential work in moving a current element \(Id\bold{L}\) through a distance \(d\bold{l}\) in a magnetic field \(\bold{B}\) is the negative of that done in moving the element \(Id\bold{l}\) through a distance \(d\bold{L}\) in the same field.

Key concepts

Magnetic field

Magnetic fields are an essential concept in electromagnetism. They are regioned around a magnetic material or a moving electric charge in which magnetic forces act. A magnetic field is typically represented by the symbol \( \mathbf{B} \) and is measured in Tesla units.

When a magnetic field interacts with an electric current or a magnetic moment, it exerts a force that can cause motion, which is often described by the vector cross product. This interaction is critical in various applications, such as electric motors and generators, where magnetic fields are used to convert electrical energy into mechanical work or vice versa. Understanding the properties of magnetic fields allows us to harness these forces effectively and is crucial for a variety of technological advancements.

Differential work

In physics, work is defined as the transfer of energy via a force causing an object to move. Differential work focuses on the very small or infinitesimal amount of work done when an object is displaced by a tiny distance. This concept is particularly useful when dealing with continuous functions or in the case of electromagnetic phenomena.

When a small current element, represented as \( I d \mathbf{L} \), moves through a magnetic field \( \mathbf{B} \), work is done as a result of the force caused by this field. The differential work can be calculated using the dot product, which mathematically expresses how much of the force contributes to moving the object along its path.

By computing this differential work, we gain insights into the subtle interactions between electromagnetic fields and moving charges or currents, which is fundamental for the design and analysis of electrical circuits and devices.

Cross product

The cross product is a vector operation used in three-dimensional mathematics and physics. It involves two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), and generates a third vector \( \mathbf{c} \) that is perpendicular to both and lies in the plane defined by them. The magnitude of \( \mathbf{c} \) is given by \( |\mathbf{a}| |\mathbf{b}| \sin \theta \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).

In electromagnetism, the cross product is used to determine the force exerted by a magnetic field on a current-carrying wire. This force is calculated by \( \mathbf{F} = I(\Delta \bold{l} \times \mathbf{B}) \), where \( I \) is the current, \( \Delta \bold{l} \) is a vector representing a segment of the wire, and \( \mathbf{B} \) is the magnetic field.

Understanding the cross product helps us predict the direction and magnitude of forces in magnetic fields, serving as a cornerstone in mechanical and electrical engineering.

Dot product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics, producing a single scalar quantity from two vectors. Given vectors \( \mathbf{a} \) and \( \mathbf{b} \), their dot product is \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \), where \( \theta \) is the angle between the vectors.

The dot product is especially useful in physics when calculating work, because it effectively measures the component of a force vector that moves an object along a displacement vector. This can be seen in electromagnetism when determining the differential work done by a magnetic field on a current-carrying wire, such as \( \bold{W} = \mathbf{F} \cdot \Delta \mathbf{l} \).

Understanding the dot product allows us to quantify how much of a vector quantity aligns or opposes another, offering crucial insights when analyzing forces and energy in physical systems.