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