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##### Top Physics 102 Electricity and Magnetism Educators    ##### Aspen F.

University of Sheffield

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So we're told that we have a spacecraft orbiting a pole star star. Okay. And pulse are stars have very fast rotation and a very strong magnetic field because of this fast rotation. Okay. And so it's in circular orbit, um, in a radius here of three times 10 to the four kilometers or three times 10 to the seventh meters, The mass of the pulse. Our capital M is 2.0 times 10 to the 30 kilograms, and we're told the magnetic field is 100 Tesla. The spacecrafts length is 0.2 kilometers or 200 meters, and its radius is 2000.4 kilometers or 40 meters on, and it moves counterclockwise in the X Y plane around the pole star and the magnetic field is in the Z direction. So this is gonna be important to remember this. So for party, it says, what is the speed of the spacecraft? Okay, well, if it's in orbit here, then, uh, we can use the condition for a particle to remain in circular orbit and the lengths law to determine. Excuse me. We can use the condition for a particle to remain in orbit to find the speed of the spacecraft. So for the particle to remain in orbit around the pole star, uh, the centripetal force of the spacecraft is going to be equal to the gravitational pole or the gravitational force of the pole star. So we can say force from centripetal F C is equal to force of gravity. We'll call that F g. Okay, Well, centripetal force is equal to the massive. The spacecraft call that little him time to the velocity squared of the spacecraft divided by our and the gravity at this distance is equal to the gravitational constant. We call that biggie right that big g to make it a little nicer looking okay, times the mass of the craft, which was a little m times the mass. The pulse are begin all over R squared. Okay, so it's solving here for gravity. The mass of the spacecraft will cancel or excuse me solving here for velocity. We find that the velocity is equal to the square root of gravitational constant biggie which is 6.67 times 10 to the minus 11 on. You can look that up if you want to know more about it. Times capital in, which is the mass of the planet divided by the distance away. The spacecraft is orbiting our plugging those values of this expression we find this is equal to 2.1 times 10 to the six. And of course, the units here are meters per second. We can go ahead and box that it is our solution for part a. Okay, start a new page here for part B, where, in part B. It says if the magnetic field points in the positive ZY direction is the IMF induced from back to front of the spacecraft or from side to side. And remember, the spacecraft moves counterclockwise in the X Y plane around the pole star. Okay, so how are we gonna do this? Well, for starters, we can consider the fact that the magnitude of the force by a magnetic field is expressed as the following. The force is equal to the charge Q multiplied by velocity V cross the magnetic field. Be so the dik. Um so here the magnetic field points in the Z direction again, right? And the velocity is in the X y plane. So since the magnetic field and the circuit is pointed along the positive, the direction and the velocity velocity is pointed along the Y access from the right hand rule. The direction of the force will be towards either right or left. So the charges will get accumulated in the direction of that force and the M F changes from side to side for the spacecraft So we can go ahead and maybe say using right hand rule. Um, how are we going to say this? We can say using the right hand rule, the E M f is induced side to side. Well, go ahead and box it in. Is our solution for B Okay, Now, Part C asked us to calculate the magnitude of this induced e enough. Well, the induce e m f is equal to the magnetic field strength be times l here. But remember, this is side beside now. Not front to back. The original elf would go back to page one. The 200 meters is front to back of the spacecraft. This is the distant side to side. So maybe we should call this l prime. It's a different l times the velocity of which the spacecraft is moving where l prime since it side to side is going to be twice the radius of the spacecraft to R and the radius was 40 meters. So this is going to be 80 meters and the velocity is the velocity we calculated in part A And of course we were told the magnetic field on the question. So playing these values in this expression, we find that this induced the IMF is 1.7 times 10 to the 10 VOCs. It's a very large induced ian meth making box set in is your solution for seat Okay, get one more page here for, uh, part D where in part D. We are asked. Describe the hazards for astronauts inside any spacecraft moving in the vicinity of a pole star. Okay, well, since we just found an induced E m F 1.7 times 10 to the 10 volts, it's very large. And due to the motion of the spacecraft in amount of magnetic field, it's getting this large induced M F on the surface of the craft, so this charge will be equally distributed throughout the surface. If some fault occurs in the insulation or something like that, any direct contact by an astronaut with the surface of the spacecraft would result in this charge being released on them and then getting electric electrocuted. So maybe we should say the large image on the surface of the craft I could Electro Cube astronauts get when you're lying here to write the astronauts well, go ahead and box set in is your solution for Part D is the potential hazard to the astronomers for the astronauts. University of Kansas
##### Top Physics 102 Electricity and Magnetism Educators    ##### Aspen F.

University of Sheffield