## College Physics (4th Edition)

The maximum torque that the motor can deliver is $1.26\times 10^{-3}~N~m$
We can find the maximum torque: $\tau_{max} = N~I~A~B~sin~\theta$ $\tau_{max} = N~I~\pi~r^2~B~sin~\theta$ $\tau_{max} = (100)~(50.0\times 10^{-3}~A)~(\pi)~(0.020~m)^2~(0.20~T)~sin~\theta$ $\tau_{max} = (1.26\times 10^{-3}~sin~\theta)~N~m$ The maximum torque that the motor can deliver is $1.26\times 10^{-3}~N~m$.