- #1
Gyroscope
Homework Statement
Would someone, please, show me that, if there is no current inside, that:
\oint \vec B \cdot d\vec l=0
Please. Thanks.
Understanding Ampère's Law: How to Solve for Zero Current in a Closed Loop
- Thread starter Gyroscope
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- Ampere's law Law
AI Thread Summary
To demonstrate that the line integral of the magnetic field B around a closed loop is zero when there is no current, Stokes's theorem can be applied to convert the line integral into a surface integral of the curl of B. By substituting Maxwell's equation for curl(B), it becomes clear that if the current density J and the displacement current dD/dt are both zero, the integral evaluates to zero. The discussion highlights the importance of understanding Stokes's theorem in this context. The user expresses a lack of familiarity with Stokes's theorem, indicating a need for foundational knowledge in electromagnetism. Mastering these concepts is essential for success in physics competitions like the IPHO.
Would someone, please, show me that, if there is no current inside, that:
\oint \vec B \cdot d\vec l=0
Please. Thanks.
- #2
marcusl
Science Advisor
- 2,958
- 668
You can do this yourself! First use Stokes's theorem to relate your integral to a surface integral over curl(B), then substitute Maxwell's eqn. for curl(B). Note what happens if J and dD/dt are zero.
- #3
Gyroscope
I do not know Stokes Theorem. :) I am preparing to IPHO.
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Figure 1 Overall Structure Diagram
Figure 2: Top view of the piston when it is cylindrical
A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N.
Figure 3: Modifying the structure to incorporate a fixed internal piston
When I modify the piston...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
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