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QUESTION
A DC motor operates by driving a current through a loop that is immersed in an external magnetic field. The interaction between the current and the external field creates a torque on the loop that causes it to rotate. A very simplified model of a DC motor is shown below. 
 
Show that if the loop is a square of side length l = 5 [cm], the current is 𝐼0 = 600.0 [mA], and the external magnetic field strength is |H ̅ | = 5[kA/m], the maximum torque experienced by the loop is [𝐼0l^2μ0𝐻0] [Nm]. 

MY ANSWER: 
As shown in Lecture 6 Slide 3, the magnetic field H ̅, and magnetic flux density B ̅, are related as follows, where μ_0 is the permittivity of free space.
B ̅=μ_0 H ̅
As shown in Lecture 5 Slide 8, in the presence of an external magnetic field, H ̅, a force is generated. If the section of wire where the force to be calculated is straight, and B ̅ is uniform then, as seen on this online source, we can define the force as
F ⃗=I_0 l ⃗×B ̅
F ⃗=I_0 l ⃗×μ_0 H ̅_0
F=I_0 lμ_0 H_0  sin⁡(θ)
Where l is the length of the wire. 
Similarly, as seen in Lecture 5 Slide 8, the torque (i.e. the turning moment of a force about an axis), is defined as follows
τ ⃗=r ⃗×F ⃗
τ ⃗=r ⃗×(I_0 l ⃗×μ_0 H ̅_0 )
τ ⃗=r ⃗×(I_0 lμ_0 H_0  sin⁡(θ) )
τ=〖rI〗_0 lμ_0 H_0  sin^2⁡(θ)

Where r is the radius of the object being rotated. 

MY QUESTIONS FOR YOU
are there any mistakes in my answer so far? I am confused about how l and r related? have I defined these correctly? because I need them to actually be equal in order to get the equation in need to reach (shown in the question stem)

Your approach is generally on the right track, but there are a few points of clarification to address regarding the relation between the side length \( l \) and the torque radius \( r \), as well as a few minor adjustments in the derivation to reach the desired result.

Let's analyze your steps closely.

### Understanding the Torque Formula
- You correctly started by relating the magnetic flux density \( \mathbf{B} \) to the magnetic field \( \mathbf{H} \) using:
  \[
  \mathbf{B} = \mu_0 \mathbf{H}
  \]
  where \( \mu_0 \) is the permeability of free space (not permittivity, which refers to \( \epsilon_0 \)).

- The force on a current-carrying wire in a magnetic field is given by:
  \[
  \mathbf{F} = I_0 \mathbf{l} \times \mathbf{B}
  \]
  or equivalently:
  \[
  F = I_0 l \mu_0 H_0 \sin(\theta)
  \]
  where \( \theta \) is the angle between the current element and the magnetic field.

### Torque Expression
Torque \( \mathbf{\tau} \) is given by:
\[
\mathbf{\tau} = \mathbf{r} \times \mathbf{F}
\]
For a rectangular or square loop, the torque is maximized when the loop is oriented perpendicular to the magnetic field (i.e., \( \theta = 90^\circ \), giving \( \sin(\theta) = 1 \)).

### Key Issue: Relation Between \( l \) and \( r \)
Here’s the important part:
- \( l \) is the side length of the square loop (which is given as 5 cm or 0.05 m).
- \( r \) represents the "lever arm" or distance from the axis of rotation to the point where the force is applied. For a square loop, this is typically half the side length, i.e.:
  \[
  r = \frac{l}{2}
  \]
  This is because the torque is generated about the center of the loop, and the force acts on the sides at a distance \( \frac{l}{2} \) from the axis of rotation.

### Correcting the Torque Expression
Given the proper relation for \( r \), the torque is expressed as:
\[
\tau = r I_0 l \mu_0 H_0 \sin(\theta)
\]
Substitute \( r = \frac{l}{2} \):
\[
\tau = \frac{l}{2} I_0 l \mu_0 H_0 \sin(\theta)
\]
At maximum torque, \( \sin(\theta) = 1 \), so:
\[
\tau_{\text{max}} = \frac{l^2}{2} I_0 \mu_0 H_0
\]
This matches the form given in the problem statement up to a factor of \( \frac{1}{2} \), and since the required expression was \( I_0 l^2 \mu_0 H_0 \), this suggests that the model considers the torque contributions from multiple segments of the loop.

### Summary of Corrections:
1. \( r \) (the effective lever arm) should be \( \frac{l}{2} \), as the force acts at a distance half the side length of the square loop from the axis of rotation.
2. When calculating torque, you need to account for this factor of \( \frac{1}{2} \), leading to \( \frac{l^2}{2} I_0 \mu_0 H_0 \) for the maximum torque.

### Next Steps:
1. Check if the lecture slides mention that the loop's symmetry leads to the final form of \( I_0 l^2 \mu_0 H_0 \), omitting the factor of \( \frac{1}{2} \).
2. Review the problem's assumptions and whether the loop's geometry or number of turns affects this calculation.

Would you like a more detailed breakdown of the relationship between the geometry and torque?

### Relative Questions:
1. Why is the magnetic field \( \mathbf{B} \) proportional to \( \mu_0 \mathbf{H} \)?
2. How does changing the side length \( l \) of the loop affect the torque?
3. What is the physical meaning of the angle \( \theta \) in the torque equation?
4. How would the torque expression change for a non-square rectangular loop?
5. What role does \( \mu_0 \) (permeability of free space) play in calculating electromagnetic forces?

**Tip:** When calculating torque, always be mindful of how the radius or lever arm is defined, especially in non-circular geometries.

A DC motor operates by driving a current through a loop that is immersed in an external magnetic field. The interaction between the current and the external field creates a torque on the loop that causes it to rotate. A very simplified model of a DC motor is shown below.

Show that if the loop is a square of side length l = 5 [cm], the current is 𝐼0 = 600.0 [mA], and the external magnetic field strength is |H ̅ | = 5[kA/m], the maximum torque experienced by the loop is [𝐼0l^2μ0𝐻0] [Nm].

Learn how to calculate the maximum torque experienced by a square loop in a DC motor using key electromagnetic concepts. The problem involves finding the torque for a loop with side length l = 5 cm, current I0 = 600 mA, and an external magnetic field of H0 = 5 kA/m. Suitable for undergraduate-level physics or electrical engineering students.

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