All of the questions relate to the loop in the picture shown right. A rectangula
ID: 2029491 • Letter: A
Question
All of the questions relate to the loop in the picture shown right. A rectangular conducting loop of mass m = 0.20 kg and width w = 0.30 m falls vertically downward at constant velocity of 8.0 m/s. A partial cross section of the loop is within a horizontal uniform magnetic field that is directed into the page and is of magnitude B-3.5 T. In the figure, h represents the height of the top edge of the loop above the bottom of the region where the field exists; h decreases as the loop falls. xl x x × × x xl xBx × × × × × × × Using Lenz's Law determine the direction of the induced current through resistor R across the bottom edge of the loop as the loop falls. (Choose the best answer-3 points) (A) To the right (i.e. counter-clockwise around the loop). 1. B) To the left(lie. dockwise around the loop).Explanation / Answer
1) Ans: (B) To the left (i.e clockwise around the loop).
The loop is coming out of the field and so the flux is reducing. So, the emf induced will be such that it increases the field B. So, the field induced must go into the page. For this, the current must be clockwise.
4) The emf induced is:
e = Bwv = 3.5*0.3*8
= 8.4 V
Then the resistance is:
R = V/I
= 8.4/I
Plug in the current I to get R.