# Magnetic field due to a rotating charged hemispherical shell

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 Ïƒ C/m2.

Let us imagine a circle on the shell drawn by rotating a radius vector PC which makes an angle Î¸ with the horizontal line AB. Imagine yet another circle on the shell drawn by rotating a radius vector PD which makes an angle Î¸+dÎ¸ 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 Ï‰, 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 = Ï‰r 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 = Ï‰ x QC = Ï‰ x PC cosÎ¸ = Ï‰ acosÎ¸ [angle PCQ is also Î¸] ........(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.