Respuesta :
Answer:
The magnitude will be "[tex](1.097 \ N/m)\hat{j}[/tex]". The further explanation is given below.
Explanation:
Trying to determine what is usual for a rectangular loop plane,
⇒ [tex]\hat{n}=(Cos35^{\circ})(-\hat{i})+(Sin35^{\circ})(\hat{k})[/tex]
Magnetic moment is given as:
μ = [tex]IA\hat{n}[/tex]
On putting the values in the above formula, we get
μ = [tex](5.5A)[(0.3m)(0.4m)][(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(-\hat{k})][/tex]
= [tex]0.66 Am^2[(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(\hat{k})][/tex]
Now the external vector-shaped torque seems to be:
Magnitude,
[tex]\sigma=\hat{\mu}\times\hat{\beta}[/tex]
On putting the values in the above formula, we get
⇒ [tex]\sigma=[(0.06Am^2)(Cos35^{\circ}(-\hat{i})+Sin35^{\circ}(-\hat{k})]\times (2.9T)(-\hat{i})[/tex]
⇒ [tex]= (0.66)(2.9)Cos35^{\circ}(\hat{i}\times \hat{i})+(0.66)(2.9)Sin35^{\circ}(\hat{k}\times \hat{k})[/tex]
⇒ [tex]= 0+(1.97 \ N/m)\hat{j}[/tex]
⇒ [tex]=(1.097 \ N/m)\hat{j}[/tex]
