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2 years ago

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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.

1 answer:

lutik1710 [3]2 years ago

3 0

Answer:

$B= 7.5*10^{-15}T$

Explanation:

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

$B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}$

Our values are:

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

Replacing,

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

$B= 7.5*10^{-15}T$

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