Answer:
The number of turns in the coil is 86.
Step-by-step explanation:
Given that,
The magnitude of maximum torque produced in the motor,
![\tau=1.7* 10^(-2)\ N-m](https://img.qammunity.org/2021/formulas/physics/college/t3c1if5uj9yucvkdzth1ryjxne9f2w6x38.png)
Area of the coil,
![A=9* 10^(-4)\ m^2](https://img.qammunity.org/2021/formulas/physics/college/snyx2ugifa0qcd6h23v4bua0ynb302gera.png)
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 :
![\tau=NIAB\ sin\theta](https://img.qammunity.org/2021/formulas/physics/college/74dchoaituu38zo0l2t0evb4vukmo665ic.png)
Here,
(maximum)
![\tau=NIAB\\\\N=(\tau)/(IAB)\\\\N=(1.7* 10^(-2))/(1.1* 9* 10^(-4)* 0.2)\\\\N=85.85\\\\N=86](https://img.qammunity.org/2021/formulas/physics/college/fgcl7vjl6x2vmo9tb19jcf23df8l6i1ywc.png)
So, the number of turns in the coil is 86. Hence, this is the required solution.