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Leto [7]
1 year ago
10

does the pair of equilibria produced by a saddle-node bifurcation have to consist of one that is stable and one that is unstable

?
Physics
1 answer:
Amiraneli [1.4K]1 year ago
3 0

This bifurcation is called a saddle-node bifurcation. In it, a pair of hyperbolic equilibria, one stable and one unstable, coalesce at the bifurcation point, annihilate each other and disappear.

<h3>What is a bifurcation equilibria?</h3>
  • The mathematical study of changes in a family of curves' qualitative or topological structure, such as the integral curves of a family of vector fields or the solutions to a family of differential equations, is known as bifurcation theory.
  • A bifurcation happens when a tiny, gradual change in a system's parameter values (the bifurcation parameters) results in an abrupt, "qualitative," or topological change in the system's behavior.
  • This term is most frequently used to refer to the mathematical study of dynamical systems.
  • Both continuous systems (represented by ordinary, delay, or partial differential equations) and discrete systems can experience bifurcations (described by maps).

To learn more about bifurcation equilibria, refer to

brainly.com/question/14728055

#SPJ4

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A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
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Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

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