Question

a square coil of 0.01 metre square area is placed perpendicular to the uniform magnetic field of 10cube weber/metersquare the flux linked with the coil is

Answer 1

Answer:

The magnetic flux through the coil = $10\ Wb$.

Explanation:

Given:

Area of the square coil, $A = 0.01\ m^2$.

Strength of the magnetic field, $B = 10^3\ Wb/m^2$.

The magnetic flux linked with the coil is defined as

$\Phi = \vec B \cdot \vec A = BA\cos\theta$

where,

$\vec A$ = area vector which has direction along the normal to the plane of the coil and magnitude equal to the surface area of the coil.

$\theta$ = angle between $\vec B$ and $\vec A$.

It is given that the square coil is placed perpendicular to the magnetic field which means the angle between $\vec B$ and $\vec A$ is $0^\circ$.

Therefore the magnetic flux through the coil is

$\Phi = BA\cos 0^\circ \\=BA\\=10^3\times 0.01\\=10\ Wb.$