Question

What is the magnetic flux through the loop shown in FIGURE EX30.4?

Short answer

The net magnetic flux through the loop, ${\mathrm{\Phi }}_{\text{net}}=4\times {10}^{-2}\mathrm{T}\cdot {\mathrm{m}}^2$.

Step-by-step solution

The magnetic flux 

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.

localid="1648896326026" ${\mathrm{\Phi }}_m=BA\cos \theta$ $\left(2\right)$

The magnitude of the flux of  2T and 1T

While $B$denotes the magnetic field as well as $A$denotes a substring circumference. Magnetic fields of various types and sizes and inclinations are required. Because the magnetic field of localid="1648896643020" $2\mathrm{T}$is directed through into sheet, whose angle was $0^o$. So the magnitude of the flux of is $2\mathrm{T}$

${\mathrm{\Phi }}_1=\left(2\mathrm{T}\right)\left(0.20\mathrm{m}\right)\left(0.20\mathrm{m}\right)\cos \left(0^{\circ }\right)=8\times {10}^{-2}\mathrm{T}\cdot {\mathrm{m}}^2$

Since localid="1648896885167" $1\mathrm{T}$'s magnetic field is beyond the sheet, whose inclination is${180}^{\circ }$

${\mathrm{\Phi }}_2=\left(1\mathrm{T}\right)\left(0.20\mathrm{m}\right)\left(0.20\mathrm{m}\right)\cos \left({180}^{\circ }\right)=-4\times {10}^{-2}\mathrm{T}\cdot {\mathrm{m}}^2$

The net flux through the loop

Since the total flux through into the looping is equal to the amount of the two fluxes, the total combined flux is

${\mathrm{\Phi }}_{\mathrm{net}}={\mathrm{\Phi }}_1+{\mathrm{\Phi }}_2=\left(8\times {10}^{-2}\mathrm{T}\cdot {\mathrm{m}}^2\right)+\left(-4\times {10}^{-2}\mathrm{T}\cdot {\mathrm{m}}^2\right)$

$=4\times {10}^{-2}\mathrm{T}\cdot {\mathrm{m}}^2$