Question

a square wire loop with sides measuring 10.0 cm is placed in a 0.430 t uniform magnetic field. the loop experiences a maximum torque of 8.90 x 10^-4 n·m. what is the current in the square? o 0.207 a o 2.07 a o 207 a o 2.07 x 10^-5 a o 0.0207 a

Explanation:

Step1: Calculate the area of the square loop

The side - length of the square loop $l = 10.0\ cm=0.100\ m$. The area of a square $A = l^{2}=(0.100\ m)^{2}=0.01\ m^{2}$.

Step2: Recall the formula for the maximum torque on a current - carrying loop in a magnetic field

The formula for the maximum torque $\tau_{max}=NIAB$, where $N = 1$ (assuming a single - turn loop), $I$ is the current, $A$ is the area of the loop, and $B$ is the magnetic field strength.

Step3: Rearrange the formula to solve for the current $I$

We know that $\tau_{max}=NIAB$, and we want to find $I$. Rearranging gives $I=\frac{\tau_{max}}{NAB}$. Substituting $N = 1$, $\tau_{max}=8.90\times 10^{-4}\ N\cdot m$, $A = 0.01\ m^{2}$, and $B = 0.430\ T$ into the formula:
\[I=\frac{8.90\times 10^{-4}\ N\cdot m}{1\times0.01\ m^{2}\times0.430\ T}\]
\[I=\frac{8.90\times 10^{-4}}{4.3\times 10^{-3}}\ A\]
\[I = 0.207\ A\]

Answer:

O 0.207 A