Question

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?

Short answer

(a) The magnitude of the magnetic flux is$3.05\times {10}^{-3}\mathrm{T}.{\mathrm{m}}^2$through this circle due to a uniform magnetic field$\overrightarrow{B}=0.230\mathrm{T}$in the direction of $+Z$direction.

(b)The magnitude of the magnetic flux is$1.83\times {10}^{-3}\mathrm{T}.{\mathrm{m}}^3$through this circle due to a uniform magnetic field $\overrightarrow{B}=0.230\mathrm{T}$at an angle of from the $+Z$direction.

(c) 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 $+Z$direction.

Step-by-step solution

Definition of flux

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

Determine magnitude of the magnetic flux in direction

The magnetic flux can be calculated as

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

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.

Determine magnitude of the magnetic flux at an angle of from the 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.

Determine the magnitude of the magnetic flux in the+ydirection

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.