Answer:
The number of turns in the coil is 86.
Explanation:
Given that,
The magnitude of maximum torque produced in the motor, [tex]\tau=1.7\times 10^{-2}\ N-m[/tex]
Area of the coil, [tex]A=9\times 10^{-4}\ m^2[/tex]
Current in the coil, I = 1.1 A
Magnetic field in the coil, B = 0.2 T
We need to find the value of N i.e. number of turns in the coil. The magnitude of torque attained in the coil is given by :
[tex]\tau=NIAB\ sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex] (maximum)
[tex]\tau=NIAB\\\\N=\dfrac{\tau}{IAB}\\\\N=\dfrac{1.7\times 10^{-2}}{1.1\times 9\times 10^{-4}\times 0.2}\\\\N=85.85\\\\N=86[/tex]
So, the number of turns in the coil is 86. Hence, this is the required solution.
