Question

A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is oriented at an angle of Ø= 25° with respect to the normal to the surface. What is the magnetic flux through the surface?

Answer 1

Answer:

Magnetic flux, $\phi=2.22\times 10^{-3}\ Wb$

Explanation:

It is given that,

Magnitude of the magnetic field, B = 0.078 T

Radius of circular loop, r = 0.1 m

The field is oriented at an angle of θ = 25° with respect to the normal to the surface. The magnetic flux through the surface is given by :

$\phi=BA\ cos\theta$

$\phi=0.078\times \pi \times (0.1)^2\ cos(25)$

$\phi=0.00222\ Wb$

or

$\phi=2.22\times 10^{-3}\ Wb$

So, the magnitude of magnetic flux through the surface is $2.22\times 10^{-3}\ Wb$. Hence, this is the required solution.