As shown in the figure, a rectangular loop with a lengthof 20.0 cm and a width w

Question

As shown in the figure, a rectangular loop with a lengthof 20.0 cm and a width w of 15.0 cm has 20 turns and carries a current of 0.250 A counterclockwise around the loop when viewed from the positive x axis. A horizontal (parallel to the x-z plane) magnetic field of magnitude 0.0550 T is oriented at an angle of 65.0 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.) x-z plane (a) Find the components of the magnetic force acting on each side of the loop Find the components of the magnetic force acting on the top section of the loop FTop, x Find the components of the magnetic force acting on the bottom section of the loop FBottom, x FBottom, )Y FBottom, z

Explanation / Answer

let,

length, l=20cm

width, w=15cm

N=20

I=0.25 A

B=0.055 T and

alpa=65 degrees

a)

F_top,x = 0

F_top,y = N*B*I*l*sin(theta)

= 20*0.055*0.25*0.2*sin(90-65)

= 23.24*10^-3 N

F_top,z = 0

and

F_bottom,x = 0

F_bottom,y = N*B*I*l*sin(theta)

= 20*0.055*0.25*0.2*sin(90-65)

= 23.24*10^-3 N

F_bottom,z = 0

and

F_left,x = N*B*I*w*sin(alpa)

= 20*0.055*0.25*0.15*sin(65)

= 37.38*10^-3 N

F_left,y=0

F_left,z = N*B*I*w*sin(alpa)

= 20*0.055*0.25*0.15*sin(65)

= 37.38*10^-3 N

and

F_right,x = N*B*I*w*sin(alpa)

= 20*0.055*0.25*0.15*sin(65)

= 37.38*10^-3 N

F_right,y=0

F_right,z = N*B*I*w*sin(alpa)

=20*0.055*0.25*0.15*sin(65)

= 37.38*10^-3 N

b)

Fnet=N*B*I*L

=20*0.055*0.25*0.2

=55*10^-3 N

c)

torque=(F_net)*(w*sin(alpa))

=55*10^-3*0.15*sin(65)

=7.48*10^-3 N.m

direction is clock wise

d)

correct option is : 0 degrees

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