Question

What is the magnetic flux linkage, in units of Weber, for a coil of 360 turns and cross sectional area of 0.133 m^2 when the magnetic flux density of 1.74 T passes through the area perpendicularly?

Answer 1

Answer:

83.3 Wb

Explanation:

The magnetic flux linkage through the coil is given by:

$N\phi = BAN sin \theta$

where

B is the magnetic field strength

A is the cross sectional area

N is the number of turns in the coil

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

In this problem:

B = 1.74 T

A = 0.133 m^2

N = 360

$\theta=90^{\circ}$

Therefore, the magnetic flux linkage is

$N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb$