3) A wire loop of radius r = 2.00 mm has 80 turns, and 2.00 A...
3) A wire loop of radius r = 2.00 mm has 80 turns, and 2.00 A of current is flowing through it. What is the maximum possible value of the torque that a magnetic field B = 2.50 T can exert on the loop?
Homework Answers
We know that torque is given by:
Torque = N*i*A*B*sin 
A = pi*R^2 = pi*(2.00*10^-3)^2 = 1.26*10^-5 m^2
R = Radius of loop = 2.00 mm = 2.00*10^-3 m
B = Magnetic field = 2.50 T
For maximum torque, sin = 1
N = number of loops = 80 turns
i = current in loop = 2.00 A
So,
Max torque = 80*2.00*1.26*10^-5*2.50*1
Max torque = 5.04*10^-3 N-m
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