Magnetic Strength and Uniform Magnetic Fields Explained

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Cutting a magnet into two equal parts results in each half having half the magnetic strength of the original magnet due to the reduction in the number of contributing atoms. The magnetic strength is defined by the formula for magnetic moment, which is the product of pole strength and the distance between the poles. When the magnet is halved, both the pole strength and the distance between poles are affected, leading to a decrease in magnetic moment. A uniform magnetic field is characterized by consistent magnetic strength and direction throughout the area. Understanding these principles clarifies the relationship between magnet size and magnetic strength.
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If a magnet is cut into exactly 2 equal parts , the magnetic strength of this part is half the magnetic strength of the original magnet.



Why does it get halved. What is the formula for magnetic strength>?
What is meant by uniform magnetic field?
 
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Again, please show you've thought about the question
 
cristo said:
Again, please show you've thought about the question

magnetic moment=pole strength*distance between poles.

Is it the quation i was looking for?
 
The magnetic strength is determined by the amount of contributors (atoms with certain electronic configurations) to the magnetic field of the magnet. If the magnet is cut in half each of the halves will get approximately half of the original amount of contributors.
 
ritwik06 said:
magnetic moment=pole strength*distance between poles.

Is it the quation i was looking for?

This equation is indeed true. So when distance b/w poles gets halved so does the magnetic moment!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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