Solving Magnetic Field of Magnetized Copper Rod: Find H Inside & Outside

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To find the magnetic field intensity H inside and outside a magnetized copper rod with uniformly distributed free current I, Ampere's law is applied. Inside the rod, the current enclosed by the Amperian loop is a fraction of I, determined by the ratio of the loop's area to the rod's cross-sectional area. The magnetization of the copper, being diamagnetic, leads to bound currents that affect the overall magnetic field. For the loop outside the rod, the total current enclosed is simply I. Understanding the distribution of current is crucial for accurately calculating H in both regions.
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Homework Statement



A long copper rod of radius R has uniformly distributed free current I. Find the value of H inside and outside the rod

Homework Equations



∫Hdl=I(free enclosed)

The Attempt at a Solution



Copper is diamagnetic so the magnetization will be circumferential and opposite of B, producing a downwards bound current inside and an upwards bound current on the surface. We can use ampere's law to calculate H:

∫Hdl=I(free enclosed). The path dl is the amperian loop inside, with s<R, so

H(2πs)= I(free enclosed)

How do I find I(free enclosed)? Isn't it just I?

Thanks! If the explanation could be as explicit as possible, that would be great.
 
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You would be right if the loop you made was outside of the rod. Then the total current enclosed is I. But, if the loop is inside the rod then only a fraction of the current exists in your loop. Knowing that the current is uniform is a hint on how to calculate the fraction of current inside of your loop.
 
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