Biot–Savart law and Ampere's law

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Biot–Savart law is generally more versatile for calculating magnetic fields, but Ampere's law is simpler in cases with high symmetry, such as an infinitely long straight wire with constant current. In such scenarios, Ampere's law allows for straightforward calculations, while Biot–Savart can still be applied effectively. The choice between the two laws often depends on the geometry of the current distribution. When symmetry is present, Ampere's law is the preferred method due to its ease of use. Understanding the context of the problem is key to selecting the appropriate law for magnetic field calculations.
minifhncc
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Hello,

Just a quick question... in which cases do I use the Biot–Savart law and Ampere's law? ie. in what cases is it easier to calculate the magnetic field with the laws?

Thanks

 
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minifhncc said:
Hello,

Just a quick question... in which cases do I use the Biot–Savart law and Ampere's law? ie. in what cases is it easier to calculate the magnetic field with the laws?

Thanks

Usually Biot Savart is more useful, but there are exceptions. The famous example is the infinitely long and straight wire with constant current. Here, Ampere's law is trivial to apply and easier than Biot Savart, although Biot Savart also works well and is not difficult to apply, in that case. Basically, Ampere's law requires a certain symmetry to be useful, but when you have that symmetry, the application is trivial, and easier than Biot Savart.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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