Respuesta :
The maximum torque on the current carrying loop of the motor is 1.26 x 10⁻⁵ N-m.
The torque on a current carrying loop is given by,
T = NiABsin(A)
For the maximum torque, sin(A) = 1, so,
T = NiAB
Where,
N is the number of turns in the loops,
i is the current in the loop which is 240 x 10⁻³ A,
A is the area of the loop which is 0.85 x 10⁻⁴ m,
B is the uniform magnetic field which is given to be 0.62 T.
So, putting the values,
T = 240 x 10⁻³ x 0.85 x 10⁻⁴ x 0.62
T = 1.26 x 10⁻⁵ N-m.
So, the maximum torque on the current carrying loop of the motor is 1.26 x 10⁻⁵ N-m.
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