1. Square loop is flat in the x-y plane with sides L and D + H. The current I ru
ID: 1370848 • Letter: 1
Question
1. Square loop is flat in the x-y plane with sides L and D + H. The current I runs clockwise as shown by the white arrowheads. The magnetic field is uniform in the z direction (B ˆk). For this problem, number the legs 1 to 4 in a clockwise manner starting with the one on the left.
(a) Find F~ 1, F~ 2, F~ 3, F~ 4. (b) (5 pts) Find F~ 1234 = F~ 1 + F~ 2 + F~ 3 + F~ 4.
2. Bent loop is bent at the dotted line so that one segment is vertical in the y-z plane. To solve this problem, we’ll consider the bent loop breakdown diagram where two fictitious legs are added to create two separate loops. Notice that the two fictitious legs add up to “zero” current. For loop a, number the legs 1 to 4 in a clockwise manner starting with the one on the left. For loop b, number the legs 5 to 8 in a clockwise manner starting with the one on the top.
(a) Find forces F~ 5, F~ 6, F~ 7, F~ 8, and F~ 5678 = F~ 5 + F~ 6 + F~ 7 + F~ 8.
(b) Find magnetic dipole moments ~µA, ~µB, and ~µAB = ~µA + ~µB
(c) Find torques ~?A, ~?B, ~?AB = ~?A + ~?B (d) (5 pts) Find potential energy U = ?~µAB · B
B=Bk D ,'H LOOP B Two fictitious legs "cancel" LOOP A SQUARE LOOP BENT LOOP BENT LOOP BREAKDOWN Figure 1: Current running in loop.Explanation / Answer
(1) We will use the feming right hand rule to find force in the branches of loop
In 1 as shown in the figure since current is negative x direction and magnetic field is in positive z direction therefore put the right hand keeping the first finger in the direction of current and middle finger in the direction of the magnetic field then the force will be shown by the thumb which will be in the positive Y direction .
We know form lorentz force that
F = BIL
therefore F1 = BIL (where L is the length of the side )
Direction of F1 is toward positive Y direction
Similarly we check in the D+H side in which current is flowing in the positive Y direction
F2 = BI(D+H) and direction will be in the positive X direction
Similarly
F3 = BIL ( in the side L where current is flowing in the positive X direction )
So the direction of the force will be in the negative Y direction
when we put our right hand first finger in the direction of the current and the middle finger in the direction of the magnetic field then our thumb will show the direction of the force which will be in the negative y direction
Similarly
F4 = BI(D+H) (side in which the direction of current is negative y direction )
The force will be in the positive Y direction
Here we note that F1 = F3 in magnitude but opposite in direction
Similarly F2 = F4 in magnitude but opposite in direction
(b) When we add all the forces
F1 +F2 + F3 +F4 = 0
because F1 and F3 cancel each other and F2 & F4 cancel each other effect.
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