Question

(a) What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0-A current in a 1.60-T field? (b) What is the torque when θ is 10.9º?

Short answer

(a) The maximum torque is $360\mathrm{N}‐\mathrm{m}$.

(b) When$\mathrm{\theta }=10.9°$, the torque equals $68.1\mathrm{N}‐\mathrm{m}$.

Step-by-step solution

Definition of Torque.

Torque can be a force applied on the different current carrying loops/segments in a uniform magnetic field.

The torque can be estimated using the expression,

$\mathbf{\tau }\mathbf{=}\mathbf{NIABsin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}$………………(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, $\mathbf{\theta }$.

Finding the maximum torque(a)

The maximum torque can be obtained when$\mathrm{\theta }=90°$.

The area of the current-carrying loop is,

$$\begin{gathered} \mathrm{A}=0.18\mathrm{m}\times 0.18\mathrm{m} \\ =0.03{\mathrm{m}}^2 \end{gathered}$$

This can be used in equation (1) to determine the torque, such that,

$$\begin{gathered} {\mathrm{\pi }}_{\max }=\mathrm{NIABsin}\left(90°\right) \\ =150\times 50\mathrm{A}\times 0.03{\mathrm{m}}^2\times 1.60\mathrm{T}\times \sin \left(90°\right) \\ =360\mathrm{N}‐\mathrm{m} \end{gathered}$$

Therefore, the maximum torque is $360\mathrm{N}‐\mathrm{m}$.

Finding the torque when the angle is (b)

Substitute the given data in equation (1), to determine the torque,

$$\begin{gathered} {\mathrm{\pi }}_{\max }=\mathrm{NIABsin}\left(90°\right) \\ =150\times 50\mathrm{A}\times 0.03{\mathrm{m}}^2\times 1.60\mathrm{T}\times \sin \left(10.9°\right) \\ =68.1\mathrm{N}‐\mathrm{m} \end{gathered}$$

Hence, the value of torque is $68.1\mathrm{N}‐\mathrm{m}$.