11
$\begingroup$

Very quick question, does a magnet contain energy? The general consensus seems to be, it does not. And this is generally confirmed by the fact that it would break the first law of thermodynamics. Whatever the hell that is (joke:)

The reason I ask is because a) I'm no genius and b) because I'm perplexed. So maybe some of you smart people could help me out please.

Here's the scenario;

Now, if I took, oh I dunno, say a metal ball and lifted it say six inches. I have converted some of my man boob calories into energy that is now stored in the ball. When I release it and it drops to original level, the energy is released. Makes sense.

Now if I took the same metal ball and rolled it along the ground, it would continue to roll until the kinetic energy was depleted, through friction and stuff like that.

Now if I take the same ball and roll it along the ground, but this time with a magnet suspended 6 inches from the ground and directly in the line of movement. The magnet is strong enough to attract the ball and is therefore lifted 6 inches and sticks to the magnet. Where has that energy come from? It can't have come from me putting the magnet there, as once I put the magnet back on the ground, I have released that energy.

As I said, I'm not smart, nor educated, just been pondering this question for a couple of days.

Would be great if you could allow my brain to get back to menial tasks.

Thanks

$\endgroup$
5
$\begingroup$

The person before has put an answer which might be more complex and answer your entropy question. however to answer your magnet issue in a simpler way concentrating on your metal ball might be more fruitful for you.

What you have is 2 fields here at work both with similar properties gravity and magnetism. let me ask you this, why does your ball fall when you raise it 6 inches and let go? gravity. the earth (and conversely the ball) attract each other. Your ball wants to always fall to the lowest energy state it can and the closer the ball and the earth are to each other the lower the potential energy as you rightly said. lift the ball and you spend your energy to move it further away and give it the gravitational potential energy. Release it and that energy is converted back into kinetic energy of movement as the ball falls to the lower gravitational potential energy state ie. closer to the earth.

your magnet is much the same. the magnet attracts the ball as the earth does, except it is an electromagnetic force not a gravitational one and it is stronger than gravity. your ball wants to fall to a lower magnetic potential energy state too and this is closer to the magnet. So therefore pulling the ball away from the magnet your providing energy to move it to a higher potential energy state and it wants to be in a lower one next to the magnet, hence attraction. now this is not as intuitive as gravity is always there and your magnet is not. but really when your magnet is not there its in the higher potential state it just cant get to a lower one until the magnet is present.

combine the two together and in opposite directions ie gravity pulling the ball down and the magnet pulling it up it will go towards the strongest. why? well because it will loose more magnetic potential energy than gravitational because the magnet is stronger. so whilst you see it fly 6 inch off the table its actually in a lower energy state than being six inches closer to earth.

hope this helps :)

$\endgroup$
  • $\begingroup$ +1 The argument by similarity of magnetic and gravitational force is helpful. So, the metal ball on the table has magnetic potential energy. This is converted to kinetic energy as it leaps towards the magnet. Just as the held ball's gravitational potential energy is converted to kinetic energy as it drops. $\endgroup$ – RedGrittyBrick Jun 28 '13 at 10:52
3
$\begingroup$

If we consider a very little ball with a very strong magnet, we can consider this ball as a test particle.

The total energy $E$ of this ball could be considered as constant and is the sum of the kinetic energy, of the gravitationnal potential, and the electromagnetic potential.

$E = T + V_G + V_{EM}$

Taking a z-axis with upwards direction, and suppose that the position of the magnet axis is at $z=0$. We suppose then that the position of the ball is always negative $z<0$.

The kinetic energy is $T = \frac{1}{2} mv^2$

The gravitationnal potential is $V_G$ = $mgz$, where $m$ is the mass of the ball.

Now, the electromagnetic potential corresponds to an attractive force, so a acceptable modelization of the potential would be $V_{EM} = \frac{\lambda}{z}$. Here $\lambda$ is positive and $z$ is negative, so $V_{EM}$ is negative and is zero at infinity ($z = - \infty$), so it is an attractive force.

Let $z_i$ and $z_f$ be the initial and final position of the ball (with $z_i$ < 0 and $z_f <0$). The ball is supposed to be at rest in its initial position ($T(z_i) = 0$).

Because the energy $E$ is constant, you have : $E(z_f) = E(z_i)$, that is :

$T(z_f) + V_G(z_f) + V_{EM}(z_f) = T(z_i) + V_G(z_i) + V_{EM}(z_i)$, that is :

$\frac{1}{2}m v_f^2 + mg(z_f - z_i) = \lambda (\large \frac{1}{z_i} - \frac{1}{z_f})$

Now, suppose that $z_f > z_i$, that means that the ball has moved upwards.

It is true that the gravitationnal potential energy has increased, because : $V_{G}(z_f) - V_{G}(z_i) = mg(z_f - z_i) >0$.

It is true that the kinetic energy has increased, because :

$T(z_f) - T(z_i) = \frac{1}{2}m v_f^2 >0$.

But, at the same time, the electromagnetic energy has been decreasing, because $V_{EM}(z_f) - V_{EM}(z_i) = \lambda (\large \frac{1}{z_f} - \frac{1}{z_i}) < 0$. (with $z_i$ < 0 and $z_f <0$))

So, the fact that the ball moves upwards is compatible with the conservation of the energy.

$\endgroup$
1
$\begingroup$

Permanent magnets do have potential energy, stored in their magnetic field. That energy can be compared to the potential energy of some compressed spring. See the picture below, representing the magnetic field lines of a magnetized sphere :

enter image description here

These lines are compressed inside the magnet. They tend to repell (because field lines do have a kind of "tension", like elastic strings. It's magnetic pressure actually), but the matter in the sphere "compress" the lines like a spring. So surely there is "potential energy" in there. That energy could be calculated by integration of the magnetic field's energy density over the whole volume of space (from inside to outside).

If the magnetic pressure is strong enough (or the matter is weak in some way), the magnetic energy could be released abruptly. It's even possible to design a "magnetic grenade" that releases that energy violently!

$\endgroup$
-2
$\begingroup$

Put a magnet on the table. This will apply a force to a piece of unmagnetised iron and attract the iron acrossthe table, the magnet is doing work against the friction of the table and giving the iron velocity and thus kinetic energy. The energy expended against the friction and given in kinetic energy comes from energy originally stored in the magnet at the beginning of the process. No equations are required to explain this!

$\endgroup$
-2
$\begingroup$

Permanent magnet can work if an opposing force is reduced. Newton’s first law of motion is undeniable and applies all kinds of forces. It must also apply on magnetic force. When an electromagnet interacts with a permanent magnet, force of permanent magnet is also present in the system. In accordance with the first law motion, we must calculate magnetic forces of both of the electromagnet and permanent magnet to deduct “potential amount of resultant force”. If the same quantity of the potential resultant force does not appear as output power, we must investigate about the lost part of the potential resultant force. If a fraction of potential amount of resultant force disappears, there must be sufficient reason of the disappearance. Once the reason is explored, it must be investigated whether the elimination of the reason is possible or not. While analyzing the results of the experiments, it can easily be found that a fraction of the iron core of electromagnet always remains unsaturated. During repulsion, permanent magnet repels like pole of electromagnet and attracts the unsaturated part of the iron core simultaneously. This phenomenon may be named as “attraction during repulsion” or “remnant attraction”. This two-fold action of permanent magnet balance (cancels) a huge part of repulsion force. In other words, permanent magnet causes matching and opposing forces simultaneously that cancel each other. Since, a complete cycle of repulsion and attraction is needed for work, the severe damage of repulsion force results in heavy loss of output power of a mechanical interaction between an electromagnet and a permanent magnet. Production of opposite forces[][1] My findings:

$\endgroup$
-3
$\begingroup$

If you were holding the magnet while the ball was lifted, you would notice that the magnet feels heavier. Once the ball is stuck to the magnet, you will feel the combined weight of the two. This will be slightly less than the weight you felt while the ball was being lifted.

The energy comes from whatever structure is holding up the magnet. If the magnet were attached to a perfectly rigid structure, the energy would ultimately come from the increase in the normal force at the structure's base.

In your example, you provided the energy while the ball was lifted. You had to exert extra force on the magnet to keep it 6 inches from the ground while the ball was being lifted.

$\endgroup$
-3
$\begingroup$

Some space crafts are accelerated by turning around a planet. So, gravity is able to increase the kinetic energy of a spacecraft. It is a kind of energy creation which we have used for many years in space travelling. If we accept that magnet has the same property of gravity then magnet has to make energy despite the energy conservation law.

$\endgroup$
  • 3
    $\begingroup$ If we accept that magnet has the same property of gravity... er what? $\endgroup$ – Kyle Kanos Dec 23 '15 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.