91影视

The magnetic field lines has been the portion of magnetization which thus goes through one loop of across By This magnet is calculated on the basis:

${\mathrm{桅}}_m=BA$

The magnet is given by if the magnetism makes an equal with axis.

${\mathrm{桅}}_{\mathrm{m}}=NBA\cos 鈦�蝇t=NBA\cos 鈦�\left(2蟺ft\right)$

The rotation's pitch is represented by the sign f. Whereas a coil's diameter is d = 18 cm, the length equals $A=蟺{\left(\frac{d}{2}\right)}^2=蟺{\left(\frac{0.18\mathrm{m}}{2}\right)}^2=25.45脳{10}^{鈭�3}{\mathrm{m}}^2$

From Faradays, the mediated is the modification of the magnetization in there in the spiral and this is provided by the type$蔚=鈭�\frac{d{\mathrm{桅}}_{\mathrm{m}}}{dt}=鈭�\frac{dNBA\cos 鈦�\left(2蟺ft\right)}{dt}$

$=NBA\frac{d\cos 鈦�\left(2蟺ft\right)}{dt}$

$=2蟺fNBA(\sin 鈦�2蟺ft)$

Because the magnetosphere is diagonally to the axis of rotation, the word "perpendicular magnetic field" is used.

$\sin 鈦�2蟺ft=\sin 鈦�{90}^{鈭�}=1$. We solve equation

$B=\frac{蔚}{2蟺fNA}$

The values $蔚,f,N\text{and}A$

$B=\frac{蔚}{2蟺fNA}$

$=\frac{170\mathrm{V}}{2蟺\left(60\mathrm{Hz}\right)\left(120\right)\left(25.45脳{10}^{鈭�3}{\mathrm{m}}^2\right)}$

$=0.15\mathrm{T}$