Answer:
[tex]\phi_B = 0.216 T m^2[/tex]
Explanation:
As we know that the length of the conductor is given as
[tex]L = 2 m[/tex]
now if it is converted into a square then we have
[tex]L = 4a[/tex]
[tex]a = \frac{L}{4} = 0.5 m[/tex]
now the are of the loop will be
[tex]A = a^2 = 0.5(0.5) = 0.25 m^2[/tex]
now the magnetic flux is defined as
[tex]\phi_B = BAcos\theta[/tex]
here we know
B = 1.0 T
[tex]\theta = 30.0^o[/tex]
[tex]\phi_B = (1.0 T)(0.25 m^2)(cos30)[/tex]
[tex]\phi_B = 0.216 T m^2[/tex]
