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A very thin circular disk of radius R = 20.00 cm has charge Q = 30.00 mC uniformly distributed on its surface. The disk rotates at a constant angular velocity ω = 5.00 rad/s around the z-axis through its center. Calculate the magnitude of the magnetic field strength on the z axis at a distance d = 2.000 × 103 cm from the center. Note that d >> R.

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Answer:

[tex]B= 7.5*10^{-15}T[/tex]

Explanation:

The magnetic field strenght on the z-axis at a distance d from the center is,

[tex]B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}[/tex]

Our values are:

[tex]R=20cm\\Q=30mc\\w=5rad/s\\d=2*10^3cm=20m[/tex]

Replacing,

[tex]B= \frac{(4\pi*10^{}-7)(30*10^{-3})(5)(0.2)^2}{8\pi(20)}[/tex]

[tex]B= 7.5*10^{-15}T[/tex]

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