A proton orbits a long charged wire, making 1.0×10^6 revolutions ...

Question

A proton orbits a long charged wire, making 1.0×10^6 revolutions per second. The radius of the orbit is 1.0 cm. What is the wire's linear charge density?

Accepted Answer

Welcome back, everyone. We are making observations about an infinitely long metal bar. Now we are told that an electron orbits that bar and we are told that the electron has an orbital radius of 0.0 to 8 m. Now, we are told that it orbits at a frequency of 1. multiplied by 10 to the power of eight revolutions per one minute. And we are tasked with finding what is going to be the numerical value of the linear charge density. We're getting started here. I do wish to acknowledge the multiple choice answers on the left hand side of our screen, those are gonna be the values in which we strive for. So without further ado let us begin. Well, what do we know? Well, given that our metal bar is infinitely long, we can treat or use the formula for the electric field due to an infinite line of charge. That formula is as follows that the magnitude of the electric field is equal to two multiplied by our desired linear charge density divided by four I multiplied by epsilon not multiplied by R. Now, as we have this electron orbiting this long metal bar, this is going to result in a centripetal force. Now, the centripetal force here is going to be given by the following equation. What we are going to have is that our centripetal force is equal to Q multiplied by E. But we also know or by filling in our value for e what we get for our centripetal force is two Q multiplied by our linear charge density divided by four multiplied by pi multiplied by epsilon, not, not multiplied by R. Now, what I'm going to do is divide both sides by cube and which leaves us with just the magnitude of the electric field. On the left hand side, we also know however that since it's a centripetal force according to Newton's second law, we have that the centripetal force is also equal to MV squared over R where B is equal to two multiplied by pi multiplied by R multiplied by our frequency. Making this substitution into our equation. We now have two Q multiplied by the linear charge density divided by four pi multiplied by epsilon knot multiplied by R multiplied by Q. This is equal to M multiplied by two multiplied by pi multiplied by R multiplied by R frequency to the power of two divided. Ir simplifying for our linear charge density. And let me go ahead and change colors here. What we get is that our linear charge density is equal to four hi multiplied by epsilon knott multiplied by the mass multiplied by four pi squared multiplied by R squared multiplied by the frequency squared divided by two Q. Now, if you remember we have that 1/ pi multiplied by epsilon knot is equal to Colo's constant. So we can simplify this down further. What we have is that our linear charge density is equal to the mass multiplied by four I squared multiplied by R squared multiplied by F to the power of two divided by a multiplied by two multiplied by Q switching colors. One more time. We are now ready to find the numerical value for our linear charged density. Using this formula. Let's go and plug in our values. We have 9.11 multiplied by 10 to the power of negative multiplied by four I squared multiplied by 0.028 to the power of two multiplied by 1.7 multiplied by 10 to the power of eight. Now that's our frequency. But we want our frequency in revolutions per second. So I'm going to divide this value by 60 but still raise it to the power of two. All of that is divided by nine multiplied by 10 to the power of nine multiplied by two, multiplied by 1.602 multiplied by 10 to the power of negative 19. What we get is 7.8 multiplied by 10 to the power of negative OMs per meter, which is equivalent to 78 PICO Kums per meter corresponding to our final answer choice of B Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

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A proton orbits a long charged wire, making 1.0×10^6 revolutions per second. The radius of the orbit is 1.0 cm. What is the wire's linear charge density?

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