Question

FIGURE EX30.5 shows a $10\mathrm{cm}\times 10\mathrm{cm}$square bent at a ${90}^{\circ }$angle. A uniform $0.050\mathrm{T}$magnetic field points downward at a ${45}^{\circ }$angle. What is the magnetic flux through the loop?

Short answer

Through the loop the magnetic flux,${\mathrm{\Phi }}_{net}=3.5\times {10}^{-4}\mathrm{T}\cdot {\mathrm{m}}^2$

Step-by-step solution

The magnetic flux 

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

localid="1648909787313" ${\mathrm{\Phi }}_m=BA$ $\left(1\right)$

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

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

The total through the bend square

Here $B$denotes the magnetic field as well as $A$denotes the substring circumference. Two fields exist, each of which has different scales and inclinations. The square consists of 2 rectangle rings of the same area and tilted points being bent at ${90}^{\circ }$. As a result, the bent square's overall flux was

${\mathrm{\Phi }}_{\text{net}}=2BA\cos \theta$ $\left(3\right)$

The magnetic flux through the loop 

Single rectangle's surface area is

$A=\left(0.10\mathrm{m}\right)\left(0.05\mathrm{m}\right)=5\times {10}^{-3}{\mathrm{m}}^2$

Then, we enter the data for $A$, $\theta$and $B$into equation $\left(3\right)$to get ${\mathrm{\Phi }}_{\text{net}}$

${\mathrm{\Phi }}_{\text{net}}=2BA\cos \theta$

$=2\left(0.050\mathrm{T}\right)\left(5\times {10}^{-3}{\mathrm{m}}^2\right)\cos \left({45}^{\circ }\right)$

$=3.5\times {10}^{-4}\mathrm{T}\cdot {\mathrm{m}}^2$