PHYS 270 M, W, F 12 pm KAssignment 1 -Due 09/0712018 Problem 27.68 Constants Par

Question

PHYS 270 M, W, F 12 pm KAssignment 1 -Due 09/0712018 Problem 27.68 Constants Part A The rectangular loop shown in the figure is pivoted about the y-axis and carries a current of 15.0A in the direction indicated. (Figure 1) If the loop is in a uniform magnetic field with magnitude 0.481T in the + x-direction, find t Express your answer using two significant figures. 3.0x10-2 N. m Correct Figure , 1of1 Part B What is the direction of the torque required to hold the loop in the position shown. 15.0 A 8.00 cm 30.0 6.00 cm

Explanation / Answer

B) Applying left hand rule of F=I(LxB) we get direction of torque is anticlockwise i.e. j^

C)

=IABsin = 15*(0.08*0.06)*0.48*sin(90-30) = -1.7*10^-2 N.m

D)

Applying left hand rule of F=I(LxB) we get direction of torque is anticlockwise i.e. –j^

E)

=IABsin = 15*(0.08*0.06)*0.48*sin(90) = 0.0N.m

F)

=IABsin = 15*(0.08*0.06)*0.48*sin(90) = 0.0 N.m

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