Cross Product Fun: Deriving Statement 2 from Statement 1

In summary, the first statement is expanded using the triple product rule to obtain the second statement, which represents the tangential component of the magnetic field subtracted from the normal component. However, when plugging in numbers, using only statement (3) leads to a different result. The book provides the second statement without explanation, but suggests using the triple product rule with the tangential component.
  • #1
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Homework Statement



1 [tex](H_1 - H_2) \times \hat n = J_s[/tex]
2 [tex](H_1t - H_2t) = \hat n \times J_s[/tex]

How does one get from the first statement to the second statement?

3 [tex]H_{1t} - H_{2t} = J_s[/tex]

This is causing an interesting problem when I plug in numbers, I would have just left it with statement (3) but if I do I get something completely different. My book gives (2) but doesn't explain how it derives it.
 
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  • #2
If the 't' means 'tangential' part use ax(bxc)=b(a.c)-c(a.b). The H part becomes H minus the normal component of H.
 
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1. What is the cross product?

The cross product is a mathematical operation that takes two vectors as input and produces a third vector that is perpendicular to both input vectors. It is often used in physics and engineering to determine the direction of a force or torque.

2. What is Statement 1 in "Cross Product Fun: Deriving Statement 2 from Statement 1"?

Statement 1 refers to the mathematical definition of the cross product, which states that the magnitude of the cross product of two vectors is equal to the product of their magnitudes multiplied by the sine of the angle between them.

3. What is Statement 2 in "Cross Product Fun: Deriving Statement 2 from Statement 1"?

Statement 2 refers to the geometric interpretation of the cross product, which states that the magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by the two vectors.

4. How do you derive Statement 2 from Statement 1?

To derive Statement 2 from Statement 1, we use the fact that the sine of the angle between two vectors is equal to the ratio of the area of the corresponding parallelogram to the product of the magnitudes of the vectors. By substituting this into the cross product definition, we can see that the magnitudes of the cross product and the area of the parallelogram are equal.

5. Why is the cross product useful in mathematics?

The cross product has many applications in mathematics, physics, and engineering. It can be used to calculate the direction of a force or torque, determine the orientation of objects in three-dimensional space, and solve problems involving vectors and their properties. It also has important applications in fields such as computer graphics, robotics, and fluid mechanics.

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