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Prove that the average magnetic field, over a sphere of radius R, due to steady currents within the sphere, is

μο 2m
(B)= ---- ----
4π R³
where m is the total dipole moment of the sphere. Show that the average magnetic field due to steady currents outside the sphere is the same as the field they produce at the center.

User Algoquant
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1 Answer

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Final answer:

The question asks for a proof of the average magnetic field within and outside a sphere caused by steady currents, involving the total magnetic dipole moment. The solution uses principles such as the Biot-Savart Law, Gauss's law for magnetism, and magnetic moments.

Step-by-step explanation:

The student's question relates to the calculation of the average magnetic field produced by steady currents within and outside a spherical volume. To prove the average magnetic field over a sphere, due to steady currents within the sphere, is μ₀ 2m / (4π R³), where m is the total magnetic dipole moment and R is the sphere's radius, we utilize the concepts of magnetostatics, specifically using the Biot-Savart Law and magnetic moments.

For currents within the sphere, the formula indicates that the average magnetic field is directly proportional to the total magnetic dipole moment of the internal currents. This calculation often involves integrating the magnetic field contribution from each current element over the sphere's volume and then averaging the result.

Regarding the magnetic field due to currents outside the sphere, if the currents are steady, the field they produce at the center of the sphere will be equivalent to the average field over the sphere's surface. This result comes from the application of Gauss's law for magnetism, which states that the net magnetic flux through a closed surface is zero, leading to the conclusion that the average field inside is equivalent to the field at the center when external currents are considered.

The calculation of the magnetic field in such configurations often involves using Ampère's Law and the principle of superposition, depending on the geometry and distribution of the steady currents.

User TatsuDn
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