As shown in the figure, a rectangular loop with a length of 20.0 cm and a width
ID: 1400054 • Letter: A
Question
As shown in the figure, a rectangular loop with a length of 20.0 cm and a width w of 15.0 cm has 20 turns and carries a current of 0.15 A counterclockwise around the loop when viewed from the positive x-axis. A horizontal (parallel to the x-z plane) magnetic field of magnitude 0.045 T is oriented at an angle of 65° relative to the perpendicular to the loop (the positive x-axis). (Assume the length and width are measured along the z and y-axes, respectively.)
(a) Find the components of the magnetic force acting on each side of the loop.
Components of the magnetic force acting on the top section of the loop.
Components of the magnetic force acting on the bottom section of the loop.
Components of the magnetic force acting on the left section of the loop.
Components of the magnetic force acting on the right section of the loop.
(b) Find the net magnetic force on the loop. (Enter the magnitude only.)
FNet =
N
(c) Find the magnetic torque on the loop.
(d) If the loop can rotate about the y-axis with only a small amount of friction, what will be the final angle of the perpendicular to the coil with respect to the direction of the magnetic field?
0°90° 180°
FTop, x = N FTop, y = N FTop, z = NExplanation / Answer
length,L = 20 cm = 0.2 m
width,w = 15 cm= 0.15 m
N= no of turns = 20
I= current = 0.15 A
B= magnetic field = 0.045 T
alpha = 65 degree
beta = 90-65 = 25 degree
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a) F_top,x = 0
F_top,y = N*I*L*B*sin(25)
= 20*0.15*0.2*0.045*sin25 = 11.4*10-3 N ( in +y direction)
F-top,Z = 0
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b) F_bottom , x = 0
F-bottom ,y = - N*I*L*B*sin(25)
= - 11.4*10-3 N ( in -y direction)
F_bottom,Z= 0
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c) F_left ,x = - N*I*w*B*sin(65)
= - 20*0.15*0.15*0.045*sin(65)
= - 18.35*10-3 N
F_left,y = 0
F_left,Z= N*I*w*B*cos(65)
= 20*0.15*0.15*0.045*cos(65) = 8.55*10-3 N
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d) F_right ,x= N*I*w*B*sin(65)
= 20*0.15*0.15*0.045*sin(65)
= 18.35*10-3 N
F_right , y= 0
F-right,Z= -N*I*w*B*cos(65)
= - 20*0.15*0.15*0.045*cos(65) = - 8.55*10-3 N
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e) F_net ( net magnetic force) = 0 ( add all the components of force in top,bottom,left,right)
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f) torque = N*I*L*B*w*sin(65) ( torque = force*perpendicular distance)
= 20*0.15*0.2*0.045*0.15*sin(65) = 3.67*10-3 N-m ( in clockwise from y -axis)
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