Question

A flat quare surface with side length 3.40cm is in the xy plane at z = 0. Calculate the magnitude of the flux through this surface produced by a magnetic field$\overrightarrow{\mathbf{B}}\mathbf{=}\left(0.200T\right)\hat{\mathbf{i}}\left(0.300T\right)\hat{\mathbf{j}}\mathbf{-}\left(0.500T\right)\hat{\mathbf{k}}$ .

Short answer

The magnitude of the flux through the square surface$\mathrm{f}=-5.78\times {10}^4\mathrm{T}.{\mathrm{m}}^2$

Step-by-step solution

Definition of the flux

The term flux may be defined as the specified physical medium having presence of force field.

Determine the magnitude of the fluxThe magnetic flux calculated as

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}}=\int \overrightarrow{\mathrm{B}}\times \mathrm{d}\overrightarrow{\mathrm{A}} \\ {\mathrm{f}}_{\mathrm{B}}=\int \left[\left(0.200\mathrm{T}\right)\hat{\mathrm{i}}+\left(0.300\mathrm{T}\right)\hat{\mathrm{j}}-\left(0.500\right)\hat{\mathrm{k}}\right]\times \mathrm{dA}\hat{\mathrm{k}} \\ {\mathrm{f}}_{\mathrm{B}}=\int \left(-0.500\mathrm{T}\right)\mathrm{dA} \\ {\mathrm{f}}_{\mathrm{B}}=\int \left(-0.500\mathrm{T}\right)\mathrm{A} \\ {\mathrm{f}}_{\mathrm{B}}=\int \left(-0.500\mathrm{T}\right){\mathrm{I}}^2 \\ {\mathrm{f}}_{\mathrm{B}}=\int \left(-0.500\mathrm{T}\right){\left(6.50\times {10}^{-2}\mathrm{m}\right)}^2 \\ {\mathrm{f}}_{\mathrm{B}}=-5.78\times {10}^{-4}\mathrm{T}.{\mathrm{m}}^2 \end{gathered}$$

Hence,the magnitude of the flux through the square surface${\mathrm{f}}_{\mathrm{B}}=-5.78\times {10}^{-4}\mathrm{T}.{\mathrm{m}}^2$