If "r\u2265R" , then the entire intensity vector flow will pass through the side surface of the cylinder, since the flow through both bases is zero. The cylinder side surface area formula is written as:
"2\\times \u03c0\\times r\\times l" . We will apply Gauss's law to a magnetic flux and we will receive:
"F=E\\times 2\\times\u03c0\\times r\\times l" , and as "E=\\frac{q}{4 \\times \u03c0 \\times \u03b5_0 \\times r^2}" , therefore "F=\\frac {\u03c1_L\\times l}{\u03b5_0}"
said expression A is a charge of the length of the cylinder. You can then write:
"E= \\frac {\u03c1_L} {2\\times \u03c0 \\times \u03b5_0 \\times r}" .
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