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marusya05 [52]
3 years ago
9

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.
Physics
1 answer:
lutik1710 [3]3 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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A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's mo
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The Young modulus E is given by:
E= \frac{F L_0}{A \Delta L}
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A is the cross-sectional area perpendicular to the force applied
L_0 is the initial length of the object
\Delta L is the increase (or decrease) in length of the object.

In our problem, L_0 = 3.0 m is the initial length of the column, E=3.0 \cdot 10^{10}N/m^2 is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
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