A rectanglular loop consists of 4 turns of wire carrying a current of 3.8 A. The
ID: 1466824 • Letter: A
Question
A rectanglular loop consists of 4 turns of wire carrying a current of 3.8 A. The loop is in the x-y plane, and the direction of flow of the current is shown in the figure. The loop has dimensions a = 2 cm and b = 5 cm. Consider a uniform magnetic field of strength 3.5×10-4 T in x, y, or z directions.
1)
If the uniform field of 3.5×10-4 T is along the +x axis, find the magnitude of the torque acting on the loop and the total force on side a and b.
=
N*m
2)
Fa =
N
3)
Fb =
N
4)
If the uniform field of 3.5×10-4 T is along the +y axis, find the magnitude of the torque acting on the loop and the total force on side a and b.
=
N*m
5)
Fa =
N
6)
Fb =
N
7)
If the uniform field of 3.5×10-4 T is along the +z axis, find the magnitude of the torque acting on the loop and the total force on side a and b.
=
N*m
8)
Fa =
N
9)
Fb =
N
Explanation / Answer
1) T = N*I*A*B*sin(90)
= 4*3.8*0.001*3.5*10^-4
= 5.32*10^-6 N.m
2) Fa = N*B*I*a*sin(90)
= 4*3.5*10^-4*3.8*0.02
= 1.064*10^-4 N
3) Fb = N*B*I*b*sin(0)
= 0
4) T = N*I*A*B*sin(90)
= 4*3.8*0.001*3.5*10^-4
= 5.32*10^-6 N.m
5) Fa = N*B*I*a*sin(0)
= 0
6) Fb = N*B*I*b*sin(90)
= 4*3.5*10^-4*3.8*0.05
= 2.66*10^-4 N
7) T = N*I*A*B*sin(0)
= 0
8) Fa = N*B*I*a*sin(90)
= 4*3.5*10^-4*3.8*0.02
= 1.064*10^-4 N
9) Fb = N*B*I*b*sin(90)
= 4*3.5*10^-4*3.8*0.05
= 2.66*10^-4 N