1) If the uniform field of 3.5×10 -4 T is along the +x axis , find the magnitude

Question

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. ? = ?

2) Fa = ?

3) Fb = ?

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. ? =?

5)Fa = ?

6)Fb =

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. ? =?

8)Fa =?

9)Fb = ?

direction of current A rectanglular loop consists of 6 turns of wire carrying a current of 2.2 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 = 6 cm. Consider a uniform magnetic field of strength 3.5x104 T in x, y or z directions.

Explanation / Answer

1) torque T = N*B*i*A*sin(90) = 6*3.5*10^-4*2.2*(0.02*0.06)*1 = 5.544*10^-6 N-m

2) Fa = B*i*L*sin(90) = 3.5*10^-4*2.2*0.02*sin(90) = 1.54*10^-5 N

3) Fb = B*i*L*sin(0) = 0 N

4) torque T = N*B*i*A*sin(90) = 6*3.5*10^-4*2.2*(0.02*0.06)*1 = 5.544*10^-6 N-m

5) Fa = B*i*L*sin(0) = 0 N

6) Fb = B*i*L*sin(90) = 3.5*10^-4*2.2*0.06*sin(90) = 4.62*10^-5 N

7) torque T = N*B*i*A*sin(90) = 6*3.5*10^-4*2.2*(0.02*0.06)*1 = 5.544*10^-6 N-m

8) Fa = B*I*a = 3.5*10^-4*2.2*0.02 = 1.54*10^-5 N

9) Fb = B*i*b = 3.5*10^-4*2.2*0.06 = 4.62*10^-5 N

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