Answer:
The magnetic flux is 0.0725 Wb
Solution:
Area of the wirer, [tex]A_{w} = 0.27 m^{2}[/tex]
Magnetic field associated with the wire, B = 0.38 T
Angle with the magnetic field, [tex]\theta = 45^{\circ}[/tex]
Now, we know that the magnetic flux is given by:
[tex]\phi_{w} = \vec{B}.\vec{A_{w}}[/tex]
[tex]\phi_{w} = BA_{w}cos\theta[/tex]
Now, using the above eqn:
[tex]\phi_{w} = 0.38\times 0.27cos45^{\circ}[/tex]
[tex]\phi_{w} = 0.38\times 0.27cos45^{\circ} = 0.0725 Wb[/tex]
