Figure 1 shows a rectangular loop of current in an external
Figure 1) shows a rectangular loop of current in an external magnetic field. The magnetic field is directed in the positive y direction of the rectangular xyz coordinate system. The sides 1 and 3 of the loop are parallel to the x axis. Initially the plane of the loop makes a 65 angle with the magnetic field. Treat side 4 as though it were one unbroken wire. The dashed line is parallel to the x axis
Part A
What is the direction of the magnetic force exerted on side 1 of the loop?
Part B
What is the direction of the magnetic force exerted on side 3 of the loop?
Part C
What is the direction of the magnetic force exerted on side 2 of the loop?
Part D
What is the direction of the torque on the loop about the axis shown as a dashed line.
Part E
What is the direction of the magnetic force exerted on side 4 of the loop?
Part F
What is the orientation of the loop once the system reaches equilibrium?
| positive y direction | |
| negative y direction | |
| negative z direction | |
| positive z direction | |
| negative x direction | |
| positive x direction |
Solution
A)
Magnetic force is given by :
F = I*(L (cross) B)
For side 1:
direction of L = +x direction
direction of B = +y direction
So, direction of force = +z direction <------ vector(cross) product of L and B
So, answer is : positive z direction
B)
On side 3:
direction of L = -x direction
direction of B = +y
So, direction of F = -z
So, answer is : negative z direction
C)
For side 2:
direction of L = cos(65) j - sin(65) k <------ in unit vector notation
direction of B = +k
So, direction of F = (cos(65) j - sin(65) k) (cross) (k)
= cos(65) i
As evident from the direction of F , answer is : positive x direction
D)
Net force on the whole loop is due to force F1(force on the side 1) and F2 (force on the side 2) being antiparallel to each other but acting on different lines of action
So, net torque is positive x axis <------answer
E)
Using the same rational as in earlier section:
force on side 4 = -x
So, answer is : negative x axis
F)
The loop will rotate clockwise(as seen from the front of the page)
So, the equilibrium point will be such that: The loop will be in the xz plane with side 1 located in the positive z direction relative to the dashed line.

