A circular area with a radius of$\mathbf{6}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}$lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic fieldlocalid="1655727900569" $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{230}\mathbf{}\mathbf{T}$(a) in the direction of +z direction; (b) at an angle of$\mathbf{53}\mathbf{.}\mathbf{1}\mathbf{°}$from the direction; (c) in the direction?

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

The magnetic flux can be calculated as

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}}=\int \overrightarrow{\mathrm{B}}\times \mathrm{d}\overrightarrow{\mathrm{A}} \\ {\mathrm{f}}_{\mathrm{B}}=\int \mathrm{BdA}\cos \left(\mathrm{\theta }\right) \\ {\mathrm{f}}_{\mathrm{B}}=\mathrm{BA}\cos \left(0\right) \\ {\mathrm{f}}_{\mathrm{B}}=\mathrm{BA} \\ {\mathrm{f}}_{\mathrm{B}}={\mathrm{\pi R}}^2\mathrm{B} \end{gathered}$$

Here R is the radius of circle and B is the magnetic field.

Substitute all the value in the above equation.

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}}=\mathrm{\pi }{\left(6.50\times {10}^{-2}\mathrm{m}\right)}^2\left(0.230\mathrm{T}\right) \\ {\mathrm{f}}_{\mathrm{B}}=3.{05}^{-3}\mathrm{T}.{\mathrm{m}}^2 \end{gathered}$$

Hence, the magnitude of the magnetic flux is$3.{05}^{-3}\mathrm{T}.{\mathrm{m}}^2$through this circle due to a uniform magnetic field $\overrightarrow{B}=0.230\mathrm{T}$/in the direction of direction.

The magnetic flux can be calculated as

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}}=\int \overrightarrow{\mathrm{B}}\times \mathrm{d}\overrightarrow{\mathrm{A}} \\ {\mathrm{f}}_{\mathrm{B}}=\int \mathrm{BdA}\cos \left(\mathrm{\theta }\right) \\ {\mathrm{f}}_{\mathrm{B}}=\mathrm{BA}\cos \left(53.1°\right) \\ {\mathrm{f}}_{\mathrm{B}}={\mathrm{\pi R}}^2\cos \left(53.1°\right) \end{gathered}$$

Here R is the radius of circle and B is the magnetic field.

Substitute all the value in the above equation.

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}}=\mathrm{\pi }\left(6.50\times {10}^{-2}\mathrm{m}\right)\left(0.250\mathrm{T}\right)\cos \left(53.1°\right) \\ {\mathrm{f}}_{\mathrm{B}}=1.83\times {10}^{-3}\mathrm{T}.{\mathrm{m}}^2 \end{gathered}$$

Hence, the magnitude of the magnetic flux is$1.83\times {10}^{-3}\mathrm{T}.{\mathrm{m}}^2$ through this circle due to a uniform magnetic field$\overrightarrow{B}=0.230\mathrm{T}$ at an angle of $53.1°$from the$+\mathrm{y}$ direction.

The magnetic flux can be calculated as

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}}=\int \overrightarrow{\mathrm{B}}\times \mathrm{d}\overrightarrow{\mathrm{A}} \\ {\mathrm{f}}_{\mathrm{B}}=\int \mathrm{BdA}\cos \left(\mathrm{\theta }\right) \\ {\mathrm{f}}_{\mathrm{B}}=\mathrm{BA}\cos \left(90°\right) \\ {\mathrm{f}}_{\mathrm{B}}=0 \end{gathered}$$

Hence, the magnitude of the magnetic flux is zero through this circle due to a uniform magnetic field$\overrightarrow{B}=0.230\mathrm{T}$in the $+\mathrm{y}$direction.