# A Magnetic Path?

I've noticed an amazing phenomenon while playing around with magnets, pulling them apart requires tremendous force(like with neodymium magnets). However, slidning them apart requires a lesser(greatly) amount of force. Why is that?

Diagram:

If I pull the magnets a part($+F_y$) the attraction force is quite high. While sliding them apart($\pm F_x$) it's less, only at the edge I'd feel a significant force but still less than the direct pull.

Also, when a magnet is attracted to a large ferromagnetic surface(like a fridge, or a iron table), I can virtually move the magnet around easily anywhere around the ferromagnet, the only force here is the frictional force due to the attraction:

This is making me think that it's easier to move a magnetic field, around another magnetic field. Imagine a path made of a magnetic field I can bring another magnetic field(assuming it's attracted to the object creating the path) and I can move it around, the only force of opposition is friction possibly. Like the magnet on the iron table, the ferromagnet creates it's own magnetic field to attract the magnet(two $B$ fields added up) and I can slide the magnet easily around.

Or, make a large electromagnet and bring a smaller magnet and we can slide it on the surface of the electromagnet easily, what explains this? The magnetic energy in the field?

I've noticed an amazing phenomenon while playing around with magnets, pulling them apart requires tremendous force(like with neodymium magnets). However, slidning them apart requires a lesser(greatly) amount of force. Why is that?

When you're pulling magnets apart, as soon as they separate you need to provide counteracting force of the same magnitude magnets attract each other with, to prevent the magnets from sticking to each other back again. The mere exertion of force, without doing any mechanical work, feels stressful and muscles consume chemical energy to maintain it.

When you slide magnet faces along each other, you only need to counteract friction force and the lateral force of the magnets, which are in sum usually much smaller than the full force of attraction.

• I have an additional question, imagine I recreate this phenomenon on a larger scale. A large plate conductor(slab) generating a magnetic field(with high current) acting like that ferromagnetic table. Then introduce an electromagnet with a powerful magnetic field and attract them to each other(like the magnet and the table), theoretically, it should be easy to slide the electromagnet(large one) around the plate slab correct? Neglecting friction. – AxtII Dec 6 '15 at 22:04
• Please post that as a different question, preferably with a picture showing the geometry of the system and distribution of electric current. In general, current-carrying slab will behave differently than ferromagnetic table. – Ján Lalinský Dec 6 '15 at 23:06

See in a permanent magnet the field at an axial position is approximately twice that at an equatorial position so the intensity of fields interacting is lesser giving you a lesser force to oppose.

It's given in your own diagram, $$\mu$$, the coefficient of friction, is usually less than 1, so the slidey-roundy friction force must be less than the sticky-downy force from the magnet.

Once you slide the magnet fields out of alignment they no longer attract so strongly.