Question

a) What is the maximum torque (in N·m) on a 130 turn square loop of wire 18.0 cm on a side that carries a 47.0 A current in a 1.60 T field? N·m(b) What is the torque (in N·m) when is 10.9°? N·m

Answer 1

a) 316.7 N*m

b) 59.9 N*m

Explanation:

Given:

Length of side (x) = 18 cm = 0.18 m

Number of turns in loop (N) = 130

Current (I) = 47 A

Magnetic field (B) = 1.6 T

The formula to find the torque is:

`\tau=N\cdot I\cdot A\cdot B\cdot\sin\theta`

a)

We calculate the area of the square loop like this:

`\begin{gathered} A=x^2 \\ \\ \text{ we replacing:} \\ \\ A=(0.18)^2 \\ \\ A=0.0324\text{ m}^2 \end{gathered}`

The maximum torque occurs when the angle is 90°, therefore:

`\begin{gathered} \tau_(max)=(130)(47)(0.0324)(1.6)\cdot\sin90\degree \\ \\ \tau_(max)=316.7\text{ Nm} \end{gathered}`

b)

We use the formula again with the same data but now the angle is equal to 10.9°, like this:

`\begin{gathered} \tau=\left(130\right)\left(47\right)\left(0.0324\right)\left(1.6\right)\cdot\sin10.9\degree \\ \\ \tau=59.9\text{ Nm} \end{gathered}`