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1 year ago

Is Ampère's law valid for all closed paths surrounding a conductor? Why is it not useful for calculating →B for all such paths?

1 answer:

8 0

Although not always practical, Ampere's law is true for all closed channels around a conductor. The integral can be calculated along many different paths, although it is difficult to do so. Think about a path that is circular but not coaxial with a long, straight wire that carries current. If the current in a conductor has enough symmetry to allow the line integral to be reduced to the size of B times an integral, Ampere's law can be used to calculate B.

<h3>What applications does Ampere's Law have?</h3>

Ampere's Law is applied to,

Find the magnetic induction caused by a long wire carrying current.
Analyze the magnetic field a toroid has.
Find the magnetic field produced by a long conducting cylinder that carries current.
Ascertain the conductor's magnetic field.

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Lelechka [254]

Answer:

a = 0.8 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the resulting force on a body is equal to the product of mass by acceleration, in this way we come to the following equation:

∑F = m*a

where:

F = forces applied [N]

m = mass = 1000 [kg]

a = acceleration [m/s²]

Now using Newton's second law.

$1200 - 400 = 1000*a\\800 = 1000 *a\\a=0.8[m/s^{2} ]$

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3 years ago

A particle of mass 10 g and charge 80 mC moves through a uniform magnetic ﬁeld,in a region where the free-fall acceleration is .

Alex777 [14]

Answer:

$B=1.223\frac{T\cdot m}{s}\frac{1}{v}$

Explanation:

According to the first Newton's law, when a particle moves with constant velocity, the sum of forces on it is zero. So, we have:

$F_m=F_g$

Here $F_m=qvBsin\theta$ is the magnetic force and $F_g=mg$ is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so $\theta=90^\circ$ . Replacing and solving for B:

$qvBsin(90^\circ)=mg\\B=\frac{mg}{qv}\\B=\frac{mg}{q}\frac{1}{v}\\B=\frac{(10*10^{-3}kg)(9.8\frac{m}{s^2})}{80*10^{-3}C}\frac{1}{v}\\B=1.223\frac{T\cdot m}{s}\frac{1}{v}$

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3 years ago

At what location in a circuit is the electrical potential energy the greatest

Eduardwww [97]

It's not the potential energy. It's just the potential.

It's greatest at the positive terminal of the battery or power supply.

4 0

4 years ago

Si la rapidez del sonido en el agua es de 1450 m/s. ¿Cuánto tiempo tardará en recorrer 500 m en el fondo del lago?

Masja [62]

Answer:

2.9

Explanation:

es 2 minutos con 9 segundos seria algo así (1450)÷(500)

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3 years ago

An electric generator uses __________ to convert mechanical energy into electrical energy.

Dafna11 [192]

Answer:

Devices that convert mechanical energy to electric energy. ... An electric generator is a device that changes kinetic energy to electrical energy through electromagnetic induction. ... [Figure1]. Q: What might happen to the current produced by an electric generator if the poles of the magnet kept reversing?

Explanation:

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3 years ago