Calcuating the integral E.ds for the paths given, what did i do wrong?

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In summary, figure 30-56 shows two circular regions with different magnetic fields, both decreasing at a constant rate. The task is to calculate the integral E.ds for each of the three dashed paths. The first step is to find the flux, which can be calculated by multiplying the magnetic field with the area of the region. However, since the fields are decreasing at a rate, the change in flux also needs to be taken into account. The voltage induced in a closed loop is proportional to the change in flux enclosed by the loop. Therefore, to find the voltage for each path, the change in flux needs to be calculated. The direction of the change in flux is important and can be determined by using the right hand rule. The direction
  • #1
mr_coffee
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Hello everyone I'm stuck on this:
3. [HRW7 30.P.034.] Figure 30-56 shows two circular regions R1 and R2 with radii r1 = 14.0 cm and r2 = 35.0 cm. In R1 there is a uniform magnetic field of magnitude B1 = 50.0 mT into the page and in R2 there is a uniform magnetic field B2 = 75.0 mT out of the page (ignore fringing). Both fields are decreasing at the rate of 8.10 mT/s.
Image:
http://www.webassign.net/hrw/hrw7_30-56.gif
Calculate the integral E.ds for each of the three dashed paths.

(a) path 1 wrong check mark V
(b) path 2 V
(c) path 3 wrong check mark V

I did the following:
Flux b1 = BA = (.05)(PI)(.14) = .02199;
Then i was confused on what I was supppose to do with the given information that the fields are decreasing at a rate of 8.10 mT/s, i saw an example in the book and they divided by that amount so i tried that and got:
2.71495 V which was wrong for path 1. I also tried -2.71495 V which was also wrong. What am i not doing right? Thanks.
 
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  • #2
Finding the flux is not enough. You have to be finding the CHANGE in flux. This can be due to two things. A change in the area A or the change in the magnetic field intensity B. Are you expected to know calculus?
 
  • #3
This is a constant rate (rather, it is an instantaneous rate of flux change).
No calculus is needed.
But you DO need your thumb in the direction of the negative change of B (flux). The actual B-field doesn't matter in this regard, only its CHANGE.
 
  • #4
When you say change of the b-field...LIke for the dashed path for loop 1, do i find the Magnetic field of that, then of loop 2, then subtract those two to get the change in magnetic flux? Or do i have to also think of that big outter loop as doing somthing? Because I see Loop 1 and Loop 2 are in different directions so those must subtract. But the big loop and the smaller loop 1 are in the same directinos, so do i add up those 2 loops?(big loop and loop 1). THen do I subtract Loop 2 since its in the opposite direction?
 
  • #5
I tried finding FLux B1 = B*A = 50.0E-3*PI*.14^2 = .0030788;
B2 = 75.0E-3*PI*.35^2 = .028863, subtracted B2-B1 = .025784 which was wrong
 
  • #6
mr_coffee said:
When you say change of the b-field...LIke for the dashed path for loop 1, do i find the Magnetic field of that, then of loop 2, then subtract those two to get the change in magnetic flux? Or do i have to also think of that big outter loop as doing somthing? Because I see Loop 1 and Loop 2 are in different directions so those must subtract. But the big loop and the smaller loop 1 are in the same directinos, so do i add up those 2 loops?(big loop and loop 1). THen do I subtract Loop 2 since its in the opposite direction?

When I meant change in magnetic field, I meant change in the magnetic field, not anything to do with the dashed loops, because they are dashed loops. In this question, the change in magnetic field (and thus change in flux) has been given to you. The voltage induced in any closed loop is proportional to the change in flux enclosed by the loop.
 

1. How do I calculate the integral E.ds?

To calculate the integral E.ds, you need to first determine the path or curve along which you want to calculate the integral. Then, you need to find the electric field vector E at each point along that path. Finally, you can use the equation E.ds = E*cos(theta)*ds to calculate the integral, where theta is the angle between E and the path, and ds is a small segment of the path.

2. What is the significance of calculating the integral E.ds?

The integral E.ds represents the work done by the electric field on a charged particle as it moves along a given path. This is important in understanding the behavior of charged particles in electric fields and can be used to calculate the potential difference between two points in the electric field.

3. What does it mean if my calculated integral E.ds is negative?

A negative value for the integral E.ds indicates that the electric field is doing work against the motion of the charged particle. This could happen if the particle is moving in the opposite direction of the electric field or if the electric field is decreasing in strength along the path.

4. How do I know if I did the calculation correctly?

You can check your calculation by comparing it to the expected value. If you know the electric field and the path, you can use the equation E.ds = E*cos(theta)*ds to calculate the integral and compare it to your result. Additionally, you can double-check your calculations and units to ensure they are correct.

5. What are some common mistakes when calculating the integral E.ds?

Some common mistakes include using the wrong path or not properly calculating the electric field at each point along the path. It is also important to pay attention to the units and make sure they are consistent throughout the calculation. Additionally, forgetting to include the cosine of the angle between E and the path can result in an incorrect calculation.

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