Question

A 2.0 m conductor is formed into a square and placed in the horizontal xy-plane. Magnetic field is oriented 30.0° above the horizontal with a strength of 1.0 T. What is the magnetic flux through the conductor

Answer 1

Answer:

The magnetic flux through the conductor is   $0.125\, T.m^{2}$

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 , $\Theta =(90-30)^{\circ}=60^{\circ}$

Let side of square be a

$\therefore 4a= 2.0\, m=>a=0.5\, m$

Therefore area of square is given by

$A=a^{2}=0.5^{2}\, m^{2}=0.25\, m^{2}$

Magnetic flux through the conductor is given by

$\phi =BA\cos \Theta =1.0\times 0.25\times \cos (60^{\circ})\, T.m^{2}$

=>$\phi =0.125\, T.m^{2}$

Thus the magnetic flux through the conductor is $0.125\, T.m^{2}$