Question

a square wire loop is placed in a 0.430 t uniform magnetic field. if the loop carries a current of 207 ma. the loop experiences a maximum torque of 8.90 x 10^-4 n*m. how long are the sides of the square? 10.0 cm 0.300 cm 5.00 cm 20.0 cm 4.00 cm

Explanation:

Step1: Recall torque formula

The maximum - torque formula for a current - carrying loop in a magnetic field is $\tau = NIAB$. For a single - turn loop ($N = 1$), $\tau=IAB$. The area of a square loop $A = s^{2}$, where $s$ is the side length of the square. So, $\tau=Is^{2}B$.

Step2: Rearrange the formula to solve for $s$

We can re - arrange the formula $\tau = Is^{2}B$ to get $s=\sqrt{\frac{\tau}{IB}}$.
Given $\tau = 8.90\times10^{-4}\ N\cdot m$, $I = 207\ mA=0.207\ A$, and $B = 0.430\ T$.
Substitute the values into the formula: $s=\sqrt{\frac{8.90\times 10^{-4}}{0.207\times0.430}}$.
First, calculate the denominator: $0.207\times0.430 = 0.08901$.
Then, $\frac{8.90\times 10^{-4}}{0.08901}\approx0.01$.
And $s=\sqrt{0.01}=0.1\ m = 10.0\ cm$.

Answer:

10.0 cm