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Home/ Questions/Q 1043115
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junaid ansari
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junaid ansari
Asked: October 8, 20242024-10-08T18:54:57+05:30 2024-10-08T18:54:57+05:30In: Education

Find the Gauss value for a position vector in Cartesian system from the origin to one unit in three dimensions.

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      Quearn Quearnist
      2024-10-08T18:55:09+05:30Added an answer on October 8, 2024 at 6:55 pm

      When discussing the Gauss value related to a position vector in the context of magnetic or electric fields, this term doesn’t directly correlate to a specific “Gauss value” for a position vector because Gauss’s law and values typically pertain to the flux of electric or magnetic field through a surface and not directly to the characteristics of position vectors themselves. However, if we’re looking to understand a basic magnetic or electric field strength (in teslas or gauss, respectively, for magnetic fields) at a point in space due to a position vector from the origin to a point, more context or specifics about the sources of the field and their relations to the vector would be needed.

      Without additional specifics—like the nature of the source of the magnetic or electric field, and whether you’re interested in fields generated by point charges, currents, or dipoles, or if you’re looking for an application of Gauss’s law (for electromagnetism) to a given configuration—it’s not possible to provide a numeric “Gauss value” for a position vector. Gauss’s law, in its essence for electricity, relates the electric flux through a closed surface to the charge enclosed by that surface, not directly assigning a value to a position vector.

      For a magnetic field, the strength is often measured in Gauss or Tesla, where 1 Tesla = 10,000 Gauss. But the strength of the field depends on the specifics of the magnetic source and its distance from the point of interest, rather than just the existence

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      junaid ansari
      2024-10-08T18:55:59+05:30Added an answer on October 8, 2024 at 6:55 pm

      b
      Explanation: The position vector in Cartesian system is given by R = x i + y j + z k.
      Div(R) = 1 + 1 + 1 = 3. By divergence theorem, ∫∫∫3.dV, where V is a cube with x = 0->1,
      y = 0->1 and z = 0->1. On integrating, we get 3 units

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