Why Is the Angle 90-theta in IBNpir^2sin(90-theta)?

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In summary, the problem involves finding the magnitude of a cross product, but the angle theta is measured from the y-axis instead of the conventional x-axis. This results in the use of 90-theta, which is just a rotation, to shift the measurement to the x-axis and make the conventional definitions of sin and cos valid. This is equivalent to taking the magnitude of sin(90-theta), which is equal to sin(theta).
  • #1
Dan453234
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Homework Statement


I know the answer to this problem is IBNpir^2sin(90-theta). What I don't get, is why the angle is 90-theta.

Homework Equations

The Attempt at a Solution

 

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  • #2
Would you prefer cos(theta)? They're measuring theta from the y axis, instead of from the x axis, which is convention. cos and sin are defined in the x-y plane as the angle measured from the x axis. sin(90-theta) = cos(theta)
 
  • #3
BiGyElLoWhAt said:
Would you prefer cos(theta)? They're measuring theta from the y axis, instead of from the x axis, which is convention. cos and sin are defined in the x-y plane as the angle measured from the x axis. sin(90-theta) = cos(theta)
i get that sin(90-theta)= cos(theta) however, how would I know to use either. I don't see how the cross product would produce cos(theta)
 
  • #4
You don't need to use the cross product. you're looking for the magnitude, which is |A||B|sin(theta).
However, this is with theta defined from the first term in the cross product, which is implicitly taken as the x-axis in the cross product calculation.
Here, they are taking theta as the measurement from the y axis, not the x. So you get 90-theta to shift the theta measurement from the y-axis to the x axis, and make the conventional sin and cos definitions valid.
I hope that makes sense.
 
  • #5
90-theta is just a rotation, maybe that's simpler. I feel like that last explanation was a little messy.
 
  • #6
BiGyElLoWhAt said:
90-theta is just a rotation, maybe that's simpler. I feel like that last explanation was a little messy.
I'm still a little bit confused. I think it may be because of the whole y-axis part. Generally I under understand that if you have a loop that has its magnetic moment 15 degrees from the direction of magnetic field it would be |A||B|sin(15), but here (probably because of the confusing y-axis part that I still don't really understand) its |A||B|sin(90-15) or |A||B|cos(15).
 
  • #7
But the magnetic moment is just under the x axis, and theta is not the measurement from the field to the moment, it's the measurement from the field to the plane of the loop, which is 90 degrees off of the moment. So the angle between the field and the moment is (using the plane of the loop measured from the y axis) 90 - theta. You're rotating the loop (or equivalently the coordinates) to get the angle of the moment out of the theta measurement.
 
  • #8
BiGyElLoWhAt said:
But the magnetic moment is just under the x axis, and theta is not the measurement from the field to the moment, it's the measurement from the field to the plane of the loop, which is 90 degrees off of the moment. So the angle between the field and the moment is (using the plane of the loop measured from the y axis) 90 - theta. You're rotating the loop (or equivalently the coordinates) to get the angle of the moment out of the theta measurement.
I still don't see how the angle between the field and the moment is 90-theta degrees. Regardless I appreciate the help.
 
  • #9
Well, you're looking for the magnitude.
sin(90-theta) = sin(-(theta-90)) = -sin(theta-90)
so when you take the magnitude, the negative sign goes away, and all of these are equivalent.
 
  • #10
Here:
With the loop the way it is in the picture, the angle between the moment and -B is 90 - theta. But since you're looking at the magnitude, you don't care if it's the angle between moment and B, or moment and -B. It's all the same, the only thing that changes is the direction.
 
  • #11
BiGyElLoWhAt said:
Here:
With the loop the way it is in the picture, the angle between the moment and -B is 90 - theta. But since you're looking at the magnitude, you don't care if it's the angle between moment and B, or moment and -B. It's all the same, the only thing that changes is the direction.
Ok I think I got it. If we were to replace the ring with a coin with the heads facing us (for visual purposes), your saying if you rotate the coin with the heads facing left that's 75 degrees and even though it's now technically 180 degrees away from the field, you are taking the magnitude so it's the same thing as facing the field.
 
  • #12
Yes. It's zero there, and it's zero 180 degrees from that as well. However, the angle theta, would be 90, as it's measured from the y-axis to the plane of the loop, or the coin, which is now in the x-z plane. So 90-theta = 0degrees, and sin(0) = 0.
 

What is the significance of the angle 90-theta in IBNpir^2sin(90-theta)?

The angle 90-theta in IBNpir^2sin(90-theta) is significant because it represents the angle of incidence in a right triangle. This angle is crucial in calculating the sine function, which is used to determine the relationship between the sides of a right triangle.

Why is the value of 90-theta used in this equation?

The value of 90-theta is used in this equation because it represents the complementary angle to theta, meaning that the sum of these two angles is equal to 90 degrees. In a right triangle, the complementary angle is important in determining the relationship between the sides and the angles.

How does the value of theta affect the calculation of IBNpir^2sin(90-theta)?

The value of theta directly affects the calculation of IBNpir^2sin(90-theta) because it is used in the sine function. As the value of theta changes, the angle of incidence in the right triangle changes, which in turn affects the value of the sine function.

What does IBNpir^2sin(90-theta) represent in terms of geometry?

IBNpir^2sin(90-theta) represents the area of a right triangle, where r is the radius of the inscribed circle and theta is the angle of incidence. This equation is derived from the formula for the area of a triangle, which is 1/2 * base * height. In this case, the base is represented by r and the height is represented by sin(90-theta).

How is IBNpir^2sin(90-theta) used in practical applications?

IBNpir^2sin(90-theta) is used in practical applications to calculate the area of a right triangle, which can be applied in various fields such as engineering, architecture, and physics. It can also be used to determine the relationship between the sides and angles of a right triangle, which is useful in solving real-world problems.

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