WEBVTT
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An electron beam with a radius of 0.90 millimeters has a measured current ๐ผ equals 30.00 microamps.
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What is the magnitude of the current density of the beam?
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We can name the beam radius, 0.90 millimeters, ๐.
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And weโre told the beamโs measured current value 30.00 microamps.
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Weโll name the current density of the beam capital ๐ฝ.
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To get started with our solution, we can recall that current density, ๐ฝ, is equal to ๐ผ over the area over which that current is spread.
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In our case, the current, ๐ผ, is given.
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And the area over which it spread is a circular cross section of the beam with radius ๐.
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When we recall that the area of a circle is equal to ๐ times its radius squared, we see we can replace ๐ด in our equation, so that ๐ฝ is now equal to ๐ผ over ๐๐ squared.
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Both ๐ผ and ๐ are given to us.
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So weโre ready to plug in and solve for ๐ฝ.
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When we do, weโre careful to use units of amperes for current and meters for our radius.
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When we calculate this fraction, we find that, to two significant figures, ๐ฝ is 12 amps per meter squared.
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Thatโs the current density of the beam.