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Juntao
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Sup all. I need help setting this problem up.
A right-triangular current loop has two 45° angles and a hypotenuse a = 5 cm. The loop lies in the x-y plane, with its hypotenuse parallel to the y-axis. A net current I = 0.42 A circulates in the loop in the direction shown in the figure. A spatially uniform magnetic field B = 0.3 T points in the +x direction. The +z direction is OUT of the screen.
Calculate:
1) F_Z (AB)=?
2) F_Z (BC)=?
3) F_Z (CA)=?
Where F_Z is the magnetic force in the Z direction on side AB, BC, or CA.
Now, I know that Force_B = iLxB
where L is length of side
and B is the magnetic field.
Obviously, I have to do some cross product action here.
I know so far that the answer for 1) is (.42A)(.05m)(.3T) sin theta=-.0063N.
Now, I'm not sure why that is the answer. Is it because the angle is 270 degrees, and that's why the answer is negative?
I'm stuck on the last two. I do know that the answer for 3 is either going to be the same as 2, or equal in magnitude but opposite in sign is my guess.
Any help greatly appreciated.
A right-triangular current loop has two 45° angles and a hypotenuse a = 5 cm. The loop lies in the x-y plane, with its hypotenuse parallel to the y-axis. A net current I = 0.42 A circulates in the loop in the direction shown in the figure. A spatially uniform magnetic field B = 0.3 T points in the +x direction. The +z direction is OUT of the screen.
Calculate:
1) F_Z (AB)=?
2) F_Z (BC)=?
3) F_Z (CA)=?
Where F_Z is the magnetic force in the Z direction on side AB, BC, or CA.
Now, I know that Force_B = iLxB
where L is length of side
and B is the magnetic field.
Obviously, I have to do some cross product action here.
I know so far that the answer for 1) is (.42A)(.05m)(.3T) sin theta=-.0063N.
Now, I'm not sure why that is the answer. Is it because the angle is 270 degrees, and that's why the answer is negative?
I'm stuck on the last two. I do know that the answer for 3 is either going to be the same as 2, or equal in magnitude but opposite in sign is my guess.
Any help greatly appreciated.
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