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A very long, rectangular loop of wire can slide without friction on a horizontal surface. Initially the loop has part of its area in a region of uniform magnetic field that has magnitude $B =$ 2.90 T and is perpendicular to the plane of the loop. The loop has dimensions 4.00 cm by 60.0 cm, mass 24.0 g, and resistance $R =$ 5.00 $\times$ 10$^{-3} \Omega$. The loop is initially at rest; then a constant force $F_{ext}$ = 0.180 N is applied to the loop to pull it out of the field (Fig. P29.46). (a) What is the acceleration of the loop when $v =$ 3.00 cm/s? (b) What are the loop's terminal speed and acceleration when the loop is moving at that terminal speed? (c) What is the acceleration of the loop when it is completely out of the magnetic field?

a. $a=4.14 \mathrm{m} / \mathrm{s}^{2}$

b. 0, $6.69 \mathrm{cm} / \mathrm{s}$

c. $7.50 \mathrm{m} / \mathrm{s}^{2}$

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for Partido es of our question was asked to find the acceleration of the loop when the velocity is equal to three times 10 to the minus two meters per second. So any meth induced in the left side of the bar because of the motion in the magnetic field. So this is the potential on the left side of the bar, which is given by, uh, potential. We're just gonna call it Absalon is equal to the velocity of the bar times the strength of the magnetic field times the length of the bar. No. Okay, so from that we can find the current I says it's equal to according to owns law potential divided by resistance. Therefore, the force from the magnetic field be is equal to hi times B times l, which is equal to v times B squared times l squared, divided by or okay, so now that we know the force of the magnetic field, we can just use Newton's second Law here and we're gonna have to start a new page for this. No, we have f external, which we were given minus force of the magnetic field of Sabi, which we just found is equal to mass times, acceleration, acceleration. Of course, being what we're trying to find here, acceleration then is equal to, um and external divided by the mass minus these times be swear times else weird, divided by the mass times the resistance. So plucking all those values into the equation we find that the acceleration is equal to 4.14 meters her second squared weakened box That is your solution for more a Barbie asked us to find the terminal velocity and the acceleration when the velocity is terminal while since the philosopher Terminal velocity is of lawsuit which it can accelerate gnome or in the acceleration they're a priori is known to be equal to zero. So right away we know that's true. A sinti for terminal zero. Okay, so the terminal velocity happens when the external force is equal to magnetic force. Therefore, this is equal to the terminal times be square times else weird, divided by resistance. So solving for the terminal velocity we find that this is equal to f external multiplied by the resistance divided by the excuse me divided by B squared multiplied by l squared. Playing those values into this expression. We find that this is equal to zero 0.69 Excuse me. 669 Then units here meters per second. Weaken Box this in as the second part of the solution for part b. Okay, now part. See assets to, um find the acceleration of the loop when it is completely out of the night night field, while the acceleration from the loop then when it's completely out of the magnetic field, is going to be the external cell oration acceleration. So F equals in May. So Emma is equal toe f external right, Therefore, a is equal to f external divided by him. Well, f external divided by m comes out to be 7.50 meters per second squared. We can go ahead and box and it is their solution.

University of Kansas