Answer:
The magnetic flux through the conductor is [tex]0.125\, T.m^{2}[/tex]
Explanation:
Given
Perimeter of square = 2.0 m
Magnetic field strength , B = 1.0 T at 30 degree with the horizontal
Therefore angle of magnetic field with the normal , [tex]\Theta =(90-30)^{\circ}=60^{\circ}[/tex]
Let side of square be a
[tex]\therefore 4a= 2.0\, m=>a=0.5\, m[/tex]
Therefore area of square is given by
[tex]A=a^{2}=0.5^{2}\, m^{2}=0.25\, m^{2}[/tex]
Magnetic flux through the conductor is given by
[tex]\phi =BA\cos \Theta =1.0\times 0.25\times \cos (60^{\circ})\, T.m^{2}[/tex]
=>[tex]\phi =0.125\, T.m^{2}[/tex]
Thus the magnetic flux through the conductor is [tex]0.125\, T.m^{2}[/tex]
