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    Magnetic field due to a rotating charged hemispherical shell

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    3. A uniformly charged hemispherical shell is rotating with angular speed w (omega) about its symmetry axis as shown. Use the Biot-Savart Law to find the magnetic field at the center of the sphere (point P). Begin by discussing the direction of B.

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    SOLUTION

    Figure shows the hemispherical uniformly charged shell. Let the surface charge density on the shell be σ C/m2.

    Let us imagine a circle on the shell drawn by rotating a radius vector PC which makes an angle θ with the horizontal line AB. Imagine yet another circle on the shell drawn by rotating a radius vector PD which makes an angle θ+dθ with the horizontal line AB. Now consider the region of the shell between the two circles, which can be considered as a thin ring.

    As the shell rotates about is axis with an angular speed ω, every point on the shell apart from the angular speed, also has a tangential linear speed. Although the angular speed of every point on the shell is same, the tangential speed depends upon the location of the point. In general, the tangential speed v of any point is given by :

    v = ωr where r is the perpendicular distance of the point from the axis of rotation.

    Thus, the tangential speed of any point on the thin ring mentioned above is given by :

    v = ω x QC = ω x PC cosθ = ω acosθ [angle PCQ is also θ] ........(1)

    Let us now consider the effect of charged thin ring rotating with a tangential speed v. As moving charge constitutes flow of current, the thing ring is also equivalent to a flowing current. As any point on ...

    Solution Summary

    The problem has been solved step by step, using integration method.

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