Question

What is the magnetic flux, in units of mWb (milliWeber), in a cross sectional area of 0.074 m^2 when the magnetic flux density of 0.92 T passes through the area at an angle of 52 degrees?

Answer 1

54 mWb

Step-by-step explanation:

The magnetic flux linkage through the coil is given by:

`\Phi = BA sin \theta`

where

B is the magnetic field strength

A is the cross sectional area

`\theta` is the angle between the direction of the field and the normal to the coil

In this problem:

B = 0.92 T

A = 0.074 m^2

`\theta=52^(\circ)`

Therefore, the magnetic flux linkage is

`\Phi = (0.92 T)(0.074 m^2) sin 52^(\circ)=0.054 Wb=54 mWb`