So here we know that four part A. We want to find the total resistance and the total length of the closed loop formed by the two Rady I. Plus the art would be two times the radius. Plus our times fada are times data would be equal to the length of the ark and so this would be equal to our times two plus data. And so the are of course, would be the resistance. And I thought it would be the angle of the ark and so we can find the total resistance by this would be equal to the resisted Vitti a row times the total length divided by a the cross sectional area in this case. So this would be there is a festivity times the radius multiplied by two plus Fada. So we're simply plugging in l. A and then divided by a And at this point, we can solve. So this would be 1.7 times 10 to be negative eighth hums meters and then multiplied by 0.24 meters plus rather to rather times two times data to plus data. This would be divided by the area of 1.20 times, 10 to the negative sixth meters squared and so the resistance is going to be equal. The total resistance would be equal to 3.4 times 10 to the negative. Third multiplied by two plus Fada Um And then the units here would, of course, be bums. So this would be our answer for the total resistance in terms of the angle of the ark. For part B, we know the area of the loop to be 1/2 our square times the angle, so the magnetic flux through the loop would be equal to B times A or simply 1/2 be r squared Fada on. And then we can solve so 1/2 times 0.5 point 15 zero Tesla's times fada times point 240 meters quantity squared and we find that the magnetic flux is going to be equal to 4.32 times 10 to the negative third fada. And then the units here would of course, be Webber's. So this would be the magnetic flux in terms of Fada or, in this case, the angle four part See, we know that the induced E M. F. Is going to be equal to the negative change in the magnetic flux. With respect the time, this is simply according to Faraday's law, with an end equaling one. So this would be equal to the negative derivative with respect the time of 1/2 b r squared data, this is gonna be equal to one half b r squared times the change in the angle with respect to time. Ah, according to angular, uh, Kenna Matics, we can we know that this is simply going to be equal to negative 1/2 times B r squared and then defaced the change in angular displacement with respect the time here, this is simply a kn angular velocity. So we can say lower case, Omega lower case Omega, of course, denoting the angular velocity. And so we can say that this is going to through arms law, we can find the current. This would be the induced E m f uh, divided by our and this would be equal to be r squared Omega, divided by two are and so I would be equal to two times 3.4 times 10 to the negative. Third multiplied by two plus data because we already had found the resistance. Um, in part A. So we're simply just plugging in. And then it would be the magnitude of the magnetic field times the radius squared times omega, the angular velocity. This would be equal to again two times. 3.4 times, 10 to the negative, Third times two plus data. Ah, we know that here it'll be on the for the numerator. It'll be the magnitude of the magnetic field times r squared. And then per the definition of the angular velocity, this would be the numerator would be multiplied by substituting in for the angular velocity Alfa. The angular acceleration multiplied by time, knowing that, of course, the angular velocity is equal to the angular acceleration multiplied by time. And so essentially, we can then find a derivative of the current with respect to time. And this would be equal to, uh, b R squared Alfa multiplied by four minus alfa t squared. And then this would be divided by 3.4 times 10 to the negative. Third times four plus Alfa T squared, quantity squared. Uh, we know that here the induced current is at a maximum when four minus Alfa t squared equals zero. So we know that I Max, occurs that, uh, four minus alpha t squared equals zero. Or we can say t is gonna be equal to the square root of four. Divided by Alfa The angular acceleration. At this instant, we can say that data would be equal to 1/2 times the angular acceleration times the time squared. This is simply angular, Kenna Matics. This would be equal to one over two times. Alfa T squared would simply be equal to four. Divided by Alfa Alphas. Cancel out and we have that data is equaling 2.0 radiance. So there's going to be a maximum current at data equaling 2.0 radiance. This is our answer for a part. See, now, finally, for part D we can say that when the current is Max, I'm Max at, uh, angular, velocity equaling Alfa T equaling outfit times the square root of four divided by Alfa. Or we can simply say, equaling the square root of four Alfa. Thus I max calculating for the maximum current we have B R squared Alfa, divided by two are aah! This would be B R squared multiplied by the square root of four Alfa, simply substituting in for the angular velocity divided by two times the resistance. Ah, we can then substitute, So this would be equal to B R. Squared times for Alfa, divided by two times 3.4 times 10 to the negative, third times two plus data. And then we can solve by plugging in Fada and the angular acceleration and essentially Oliver Constance. So the maximum current occurs. Rather, the value for the maximum current will be 0.150 Tesla's multiplied by 0.24 meters Quantity squared, um, multiplied by the square root of four times 12 radiance per second squared Aah! This would be divided by two times, 3.4 times 10 to the negative. Third multiplied by two plus 2.0, and we find that the maximum current is gonna be equal to 2.20 And pierce, This would be our final answer for Part D. That is the end of the solution. Thank you. For once