Question 3 (8 pts): A current-carrying loop, carrying a current- IA in the count

Question

Question 3 (8 pts): A current-carrying loop, carrying a current- IA in the counter-clockwise direction, enters a magnetie field B-10 Tesla as shown below. The magnetic field is pointing out of the page (z-direction), 20 cm 10 cm cm a) Using what you know of magnetic forces on current-carrying wires (or segments) Verify that at nt loop experiences a force in the +x direction. (Hint: no need to consider the entire loop, only the right-most segment). b) What is the value of the force F that the loop experiences in the +x direction? d When the loop is completely inside the magnetic field region, the field is switched to the ty direction without changing its magnitude. Verify that the current loop will now experience a torque causing it to rotate around the x-axis. Will it rotate clock-wise or counter-clockwise when looking from the right of the loop? What is the magnitude of the torque on the loop? d) Go back to the situation in parts a and b of this problem. What is the net force on the loop when it is fully inside the magnetic field? Will it keep traveling in the *x direction? If yes, what will happen when it starts emerging from the other side? e)

Explanation / Answer

(a) Using Fleming's left hand rule

where thumb, index and middle finger is in perpendicular form and they denotes
thumb denotes direction of force
index finger denotes direction of Magnetic field
Middle Finger denotes direction of Current

So according to this rule the direcion of force is in right side (+x direction)

(b) F = BIL = 10 x 1 x .05

F = 0.5 N

(C) When the loop is completely inside the mangnetic field and direction is towards +y

so now there is no force on small (5 cm) segment because the are in the same plane of magnetic field, so now there is only force on long segments and the direction is by fleming's left hand rule

so on upper segment direction of force is in -z direction/inside paper
on lower segment direction of force is in +z direction/out of paper

So it rotate "Clockwise" when looking from right side.

(D) F = NIAB
N = number of turns = 1
I = current = 1 A
A = area of coil = L x b = .05 x .10 = 0.005 m^2
B = 10 T

F = 1 x 1 x 0.005 x 10

F = 0.05 N

(e) When it is in fully

Net Force = 0 (Beacuse the same amount of force but opposite in direction is applied at left side small segment so it stops moving)

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