0.5
The power factor \( \cos(\phi) \) is given by: \[ \cos(\phi) = \frac{R}{Z} \] where \( Z = \sqrt{R^2 + X_C^2} \). Given \( \frac{X_C}{R} = \frac{4}{3} \), we find: \[ Z = R \sqrt{1 + \left(\frac{4}{3}\right)^2} = R \sqrt{1 + \frac{16}{9}} = R \sqrt{\frac{25}{9}} = \frac{5R}{3} \] Thus, \( \cos(\phi) = \frac{R}{\frac{5R}{3}} = 0.6 \).
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
