# Question 5c5a5

Oct 17, 2017

a) ${\Phi}_{B} = 0.0306$ Wb
b) ${\Phi}_{B} = 0.0184$ Wb

#### Explanation:

Magnetic flux is calculated as ${\Phi}_{B} = B \cdot A \cdot \cos \theta$
where $\theta$ is the angle between the direction of the field and the normal to the surface enclosed by the loop.

In this case, the field is 0.230 T, and the area is $A = \pi \cdot {r}^{2} = 0.0133 {m}^{2}$

(Note the necessary change from cm to m.)

So, if $\theta$ = 0°, as in part a of your question (If the circle lies in the $x - y$ plane, the normal points perpendicular to the plane - in the direction of the z-axis.)

Phi_B= (0.230*0.0133)*cos(0°)=0.00306 Wb (webers)

And in part b, where $\theta$= 53.1°

Phi_B= (0.230*0.0133)*cos(53.1°)=0.00184# Wb