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   25 with respect to the normal to the surface. What is the magnetic flux through the surface?

Answer 1

The magnetic flux through surface is
`2.22 * 10^(-3)` Wb

Step-by-step explanation:

Given :

Magnitude of magnetic field
`B = 0.078` T

Radius of circle
`r = 0.10` m

Angle between field and surface normal
`\theta =` 25°

From the formula of flux,

`\phi = B.A`

`\phi = BA\cos \theta`

Where
`\theta =` angle between magnetic field line and surface normal,
`A =` area of circular surface.

`A = \pi r^(2)`

`A = 3.14 * (0.10) ^(2)`

`A = 0.0314`
`m^(2)`

Magnetic flux is given by,

`\phi = 0.078 * 0.0314 * \cos 25`

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

Therefore, the magnetic flux through surface is
`2.22 * 10^(-3)` Wb