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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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Given the data in the question;

  • Mass of car; m = 200kg
  • Acceleration of the car; a = 0.5m/s^2
  • Distance covered by the car; d = 10m

Work done; W = \ ?

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

Work\ done = f * d

Where f is force applied and d is distance travelled.

To determine the work done by the man, we first solve for the force applied F.

From Newton's Second Law; Force \ F = m * a

We substitute our given values into the expression

F = m * a \\\\F = 200kg * 0.5m/s^2\\\\F = 100kg.m/s^2

Next we substitute our values into the expression of work done above.

Work \ done = f * d\\\\Work \ done = 100kg.m/s^2 * 10m\\\\Work \ done = 1000kgm^2/s^2\\\\Work \ done = 1000J

Therefore, the work done by the man pushing the car over the given distance is 1000J.

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