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.