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2 years ago

What is the net magnetic flux through any closed surface?

Physics

1 answer:

Paul [167]2 years ago

3 0

Answer:

The net magnetic flux through any closed surface must always be zero.

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Describe, using the relevant physics, how moving a magnet near a [ 1 2 ] solenoid induces a voltage across it. How does the spee

Svetllana [295]

Answer:

Explanation:

Moving a magnet might cause a change in the magnetic field going through the solenoid. Whether or not it will change depends on the movement.

According to Faraday's law of induction a voltage is induced in a coil by a change in the magnetic flux. Magnetic flux is defined as the dot product of the magnetic field (a vector field) by the area enclosed by a loop of the coil.

$\Phi B = -\int{B} \, dA$

The voltage is induced by the variation of the magnetic flux:

$\epsilon = -N * \frac{d \Phi B}{dt}$

Where

ε: electromotive fore

N: number of turns in the coil

ΦB: magnetic flux

Moving the magnet faster would increase the rare of change of the magnetic flux, resulting in higher induced voltage.

Turning the magnet upside down would invert the direction of the magnetic field, reversing the voltage induced.

5 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

2 years ago

What particles bond together to

yanalaym [24]

Answer:

atoms bond together and forms molecules

5 0

2 years ago

Read 2 more answers

A student bangs a brick at the head of a table. Three students are positioned equal distance from the head with their hands on t

iris [78.8K]

Explanation:

that the people closer too the head of the table will feel more vibrations than the people at the end of the table. since the vibrations will slow down as they travel farther down the table

Hope this helps!!

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2 years ago

The gravitational force is always attractive, while the electric force is both attractive and replusive. What accounts for this

tangare [24]

The gravitational force is gravity meanwhile the electric force is electric

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2 years ago