Honors Practice Problem #9

The torques = F per arm times distance of arm from axis of rotation x 2 arms per turn x # of turns.

Using force per arm = ILxB I get F = 2X.125 X.048 = .012

Distance to axis of rotation = 1/2 a side = .125/2

torque = .012 x .125/2 x 2 arms x 75 turns = =.00075 x 2 x75 = .12 Nm or answer C [ not D]. Check me and see if you agree.