Question

1. Consider a rectangular conducting loop of length l, width w, mass m, and resistance R. Due to gravity g, it is falling out of a uniform magnetic field that points out of the page. At the time shown in the figure, the rate at which heat is released from the loop reaches a constant value P. O © Boo O O BrŐ Figure 1: Loop falling out of a magnetic field (a) Find the magnetic field B in terms of the given variables. Follow the below steps if you get stuck! i. Given the dissipation power P, what is the current I through the loop? What must be the emf E that creates that current? ii. Use Faraday's Law to write the magnetic field B in terms of the given variables and the loop's velocity v. (By now you should have 1 equation for 2 unknowns B and v. Let's find the 2nd one in the next step.) iii. Note that if P is a constant, so is the velocity v. What does that mean for the acceleration of the loop and relation between the forces acting on the loop? Use this to find the 2nd equation for B and v. (Hint: one of the 2 forces is magnetic and you need to show its direction using Lenz's Law and the Right Hand Rule!) iv. Eliminate v and solve for B. (b) When the loop is completely in the region with no magnetic field, explain what happens to the power P, the forces acted on the loop and its velocity. (c) Bonus: How does your answer change if the lower region has another uniform magnetic field B1 < B also pointing out of the page? Here are some hints i. There will be two contributions in the change in magnetic flux when using Faraday's Law. Be careful about their signs. ii. There will be a 2nd magnetic force on the lower edge of the loop when using Newton's 2nd Law.

Answer 1

= APRIL sol (1) The urrent is 1. lk. - and induced emt, É - IR ☺ using Farciday's law. Egracity : Imago mg = I WB] - ciq.o¢ © Substitute ineg o we get. mg & WB DPB WB/R ng offweco .. aquating eq. BE ngre WTp

(b The then There is no change of magnetic fler. because Wellent is zero. Gravity is acted velocity increases.