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Semenov [28]
4 years ago
13

A magnet is placed inside a small cube which is placed inside a larger cube which has eight times the volume of the smaller cube

How does the net magnetic flux through the large cube compare to that through the smaller cube?
Physics
1 answer:
alexdok [17]4 years ago
4 0

Answer:

The magnetic flux through the two cubes is zero in both cases

Explanation:

To answer this question, we have to think about the nature of magnetic fields.

In fact, we know that magnetic sources always exist only as magnetic dipoles: this means that a magnet always has a north pole (from which the magnetic field lines go away) and a south pole (into which the magnetic field lines return). There exist no magnetic monopoles: even when we cut a magnet in a half, we end up having two magnets, each of them having its own north pole and south pole.

A direct consequence of this fact is that if we take a closed surface, such as a cube surrounding the magnet, the magnetic flux through the cube is always zero. This is because all the field lines going out the surface of the cube always return inside the cube on another point. Since the magnetic flux basically represents the number of field lines passing through the surface of the cube, this means that the net positive magnetic flux (lines going out of the cube) is equal to the net negative magnetic flux (lines going into the cube).

As a result, the magnetic flux is zero for both the smaller cube and the larger cube.

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Answer:

q_1=\pm0.03 \mu C and q_2=\pm0.02 \mu C.

Explanation:

According to Coulomb's law, the magnitude of  force between two point object having change q_1 and q_2 and by a dicstanced is

F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)

Where, \epsilon_0 is the permitivity of free space and

\frac{1}{4\pi\spsilon_0}=9\times10^9 in SI unit.

Before  dcollision:

Charges on both the sphere are q_1 and q_2, d=20cm=0.2m, and F_c=1.35\times10^{-4} N

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1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}

\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

\frac{q_1+q_2}{2}

d=20cm=0.2m, and F_c=1.406\times10^{-4} N

So, from equation (i)

1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}

\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}

\Rightarrow q_1+q_2=\pm5\times 10^{-8}

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)

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The equation (ii) become:

(5\times 10^{-8}-q_2)q_2=6\times10^{-16}

\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0

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From equation (iii)

q_1=2\times10^{-8}, 3\times10^{-8}

So, the magnitude of initial charges on both the sphere are 3\times10^{-8} Coulombs=0.03 \mu C and 2\times10^{-8} Colombs or 0.02 \mu C.

Considerion the nature of charges too,

q_1=\pm0.03 \mu C and q_2=\pm0.02 \mu C.

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Answer:

The final stored energy will become half.

Explanation:

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E=\dfrac{1}{2}CV^2

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d=Distance between plates

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E=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d}\times V^2

If the distance between plates get double ,say d' = 2 d

Then stored energy

E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d'}\times V^2

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Answer:

Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.

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