91Ó°ÊÓ

Torque is a measurement of the force required to rotate an object around an axis and be expressed as,

$\mathbf{Ï„}\mathbf{=}\mathbf{NIABsin}\mathbf{\left(}\mathbf{θ}\mathbf{\right)}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

where the number of turns, $\mathbf{N}$, the current,$\mathbf{I}$, the area of the loop,$\mathbf{A}$, and the inclination angle of the loop, $\mathbf{θ}$.

The maximum torque is obtained when$\mathrm{θ}=90°$and therefore$\sin \left(90°\right)=1$, and in that case, the equation (1) can be written as,

${\mathrm{Ï„}}_{\mathrm{mas}}=\mathrm{NIAB}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.…â¶Ä¦(2)

we will compare the torque in each case with the maximum torque to get the angle.

$$\begin{gathered} \frac{\mathrm{Ï„}}{{\mathrm{Ï„}}_{\mathrm{mas}}}=\frac{\mathrm{NIAB}\sin \left(\mathrm{θ}\right)}{\mathrm{NIAB}} \\ =\sin \left(\mathrm{θ}\right) \end{gathered}$$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(3)

Let us calculate the angle using equation (3), such that,

$$\begin{gathered} \sin \left(\mathrm{θ}\right)=\frac{90}{100} \\ \mathrm{θ}=\arcsin \left(\frac{90}{100}\right) \\ \mathrm{θ}=64° \end{gathered}$$

Hence, the required value is $64°$.

Calculate the angle using equation (3), which will give,

$$\begin{gathered} \sin \left(\mathrm{θ}\right)=\frac{50}{100} \\ \mathrm{θ}=\arcsin \left(\frac{50}{100}\right) \\ \mathrm{θ}=30° \end{gathered}$$

Hence, the required angle is $30°$.

Considering the given information and using equation (3), we will get

$$\begin{gathered} \sin \left(\mathrm{θ}\right)=\frac{10}{100} \\ \mathrm{θ}=\arcsin \left(\frac{10}{100}\right) \\ \mathrm{θ}=5.7° \end{gathered}$$

Therefore, the required angle is $5.7°$