Hey guys , i am in need of some help with a physics question as I have had a lot

Question

Hey guys , i am in need of some help with a physics question as I have had a lot of family issues going on. 1500 points! Thanks in advance.

As shown in the figure, s 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 C .39 A counterclockwise around the loop when viewed from the positive x-axis. A horizontal (parallel to the x-z plane) magnetic field of magnitude 0.051 T is oriented at an angle of 65 degree 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.) Find the components of the magnetic force acting on each side of the loop. Find the net magnetic force on the loop. (Enter the magnitude only.) Find the magnetic torque on the loop. 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?

Explanation / Answer

a)

TOP:

Fx = 0 N

Fy = N B L i sin(alpha) = 20*0.051*0.20*0.39*sin(90-65) = 3.36 x 10^-2 N

Fz = 0 N

----------------------------------------------------------------------------------------------

Bottom:

Fx = 0 N

Fy = -N B L i sin(alpha) = -20*0.051*0.20*0.39*sin(90-65) = -3.36 x 10^-2 N

Fz = 0 N

----------------------------------------------------------------------------------------------

Left:

Fx = -N B L i sin(alpha) = -20*0.051*0.15*0.39*sin(90)*sin(65) = -5.41 x 10^-2 N

Fy = 0 N

Fz = N B L i sin(alpha) = 20*0.051*0.15*0.39*sin(90)*cos(65) = 2.52 x 10^-2 N

----------------------------------------------------------------------------------------------

Right:

Fx = N B L i sin(90) sin(alpha) = 20*0.051*0.15*0.39*sin(90)*sin(65) = 5.41 x 10^-2 N

Fy = 0 N

Fz = -B L i sin(90) cos(alpha) = -0.051*0.15*0.39*sin(90)*cos(65) = -2.52 x 10^-2 N

b)

Fx_net = 0 + 0 - 5.41 + 5.41 = 0 N

Fy_net = 3.36 - 3.36 + 0 + 0 = 0 N

Fz_net = 0 + 0 + 2.52 - 2.52 = 0 N

==> F_net = 0 N

c)

torque = i N A B sin(alpha)

= i N w l B sin(alpha)

= 0.39 * 20 * 0.20 * 0.15 * 0.051 * sin(65)

= 1.08 x 10^-2 N.m

direction: -y

d)

0 degrees

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