Question

A magnetic field has a magnitude of 0.35 T and is uniform over a square loop (1 turn) 0.2 m per side. The field is oriented at an angle of (circle with line vertically through it) 35 degrees with respect to the normal of the surface. Find the flux through the loop.

Answer 1

Explanation:

It is given that,

Magnitude of magnetic field, B = 0.35 T

Side of square, a = 0.2

Area of square, $A=(0.2)^2=0.04\ m^2$

Angle between field and normal of the surface is 35 degrees. So, the angle between the magnetic field and area is, $\theta=90-35=55^{\circ}$

We need to find the flux through the loop. Magnetic flux is given by :

$\phi=BA\ cos\ \theta$

$\phi=0.35\times 0.04\ cos\ (55)$

$\phi=0.008\ T-m^2$

$\phi=0.008\ Wb$

or

$\phi=8\ mWb$

Hence, this is the required solution.