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

0.00221Tm^2

Step-by-step explanation:

B( magnitude of magnetic field )=0.078T

r=0.1m

theta=25°

magnetic flux=?

Area=πr^2

=3.14×0.1^2

=0.0314m^2

Magnetic flux= BACostheta

=0.078×0.0314×Cos25°

=0.00221Tm^2 /0.00221wb

Answer 2

Magnetic flux,
`\phi=2.22* 10^(-3)\ Wb`

Step-by-step 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* \pi * (0.1)^2\ cos(25)`

`\phi=0.00222\ Wb`

or

`\phi=2.22* 10^(-3)\ Wb`

So, the magnitude of magnetic flux through the surface is
`2.22* 10^(-3)\ Wb`. Hence, this is the required solution.