Question

3. A conductor having an area 0.14m2 is placed in the horizontal -plane. A magnetic field is oriented 30.0P with the normal to the plane of the conductor. with a strength of 1.0 T What is the magnetic flux through the conductor?

Answer 1

By the problem statement, Area of the conductor placed in the x-y plane is $A=0.14m^2$

Magnetic field is oriented $\theta = 30^0$ with the normal to the plane of the conductor.

Strentgth of the magnetic field is

$\hat{n}$ is the unit normal vector of the area 'A'. Therefore area vector can be written as $\vec{A}=A\hat{n}$

The magnetic flux through the conductor is $\phi=\vec{B}\cdot \vec{A}$

$=\vec{B}\cdot A\hat{n}=A(\vec{B}\cdot \hat{n})=ABcos \theta$

  $=1 \times 0.14\times cos30^0=0.12 (Weber)$