FIGURE EX30.8 shows a $2.0-\mathrm{cm}-$diameter solenoid passing through the center of a $6.0-\mathrm{cm}-$diameter loop. The magnetic field inside the solenoid is $0.20\mathrm{T}$. What is the magnetic flux through the loop when it is perpendicular to the solenoid and when it is tilted at a role="math" localid="1648914344990" ${60}^{\circ }$angle?

A magnetic flux $\mathrm{\Phi }$seems to be the magnitude of magnetic field which passes through some kind of ring of area $A$. If the magnetic field lies straight towards the plane's normal, This magnetic flux was provided by

${\mathrm{\Phi }}_m=BA$ $\left(1\right)$

The magnetic flux, however, would be supplied by if the magnetic field forms an angular position only with planes.

${\mathrm{\Phi }}_{\mathrm{m}}=BA\cos \theta$ $\left(2\right)$

The inclination of the parallel solenoid is perpendicular to the area's normal, hence the inclination is zero $\theta =0^{\circ }$. The flux does have a region equivalent to the solenoid's sector, as well as any magnetic field beyond this zone had zero flux. The solenoid has a diameter of $d=2\mathrm{cm}$. As a result, the area is

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{0.02\mathrm{m}}{2}\right)}^2=3.14\times {10}^{-4}{\mathrm{m}}^2$

Then, you put the data for $A$, $\theta$and $B$to the equation (2) to obtain ${\mathrm{\Phi }}_{\perp }$

role="math" localid="1648915205364" ${\mathrm{\Phi }}_{\perp }=BA\cos \theta =\left(0.20\mathrm{T}\right)\left(3.14\times {10}^{-4}{\mathrm{m}}^2\right)\cos 0^{\circ }$

role="math" localid="1648914759771" $=6.28\times {10}^{-5}\mathrm{Wb}$

The area would be tilted by the angle of ${60}^{\circ }$,and thus surface area in the another instance would be

$A=\frac{3.14\times {10}^{-4}{\mathrm{m}}^2}{\cos {60}^{\circ }}=6.28\times {10}^4{\mathrm{m}}^2$

Then, you put the data for $A$, $\theta$and $B$in the equation (2) to obtain ${60}^{\circ }$

role="math" localid="1648915352396" ${\mathrm{\Phi }}_{60}=BA\cos \theta =\left(0.20\mathrm{T}\right)\left(6.28\times {10}^{-4}{\mathrm{m}}^2\right)\cos {60}^{\circ }$

role="math" localid="1648915305484" $=6.28\times {10}^{-5}\mathrm{Wb}$