- #1
cestlavie
- 10
- 3
- Homework Statement
- A wire loop with 60 turns is formed into a square with sides of length 𝑠. The loop is in the presence of a 1.00 T uniform magnetic field 𝐵 that points in the negative 𝑦 direction. The plane of the loop is tilted off the 𝑥-axis by ##\theta=15##. If 𝑖=1.30 A of current flows through the loop and the loop experiences a torque of magnitude 0.0727 N⋅m , what are the lengths of the sides s of the square loop, in centimeters?
- Relevant Equations
- ## A=x^2##
## \tau=NIABsin\theta##
If ##\tau= 0.0727, N=60, i=1.3, B=1.0,## and ##\theta=15##, I tried the following calculation:
##\tau=NIABsin\theta##
##\tau=NIs^2Bsin\theta##
##s^2=\frac {\tau} {NIBsin\theta}=\frac {.0727} {60*1.3*1*sin(15)}=0.0632 m=6.32 cm##
The answer is probably right in front of me, but I don't know what I am doing wrong.
##\tau=NIABsin\theta##
##\tau=NIs^2Bsin\theta##
##s^2=\frac {\tau} {NIBsin\theta}=\frac {.0727} {60*1.3*1*sin(15)}=0.0632 m=6.32 cm##
The answer is probably right in front of me, but I don't know what I am doing wrong.