Question

The magnetic field $B_vec$ in a certain region is 0.128-axis in the Figure

A) What is the magnetic flux across the surface abcd in the figure?
B) What is the magnetic flux across the surface befc?
C) What is the magnetic flux across the surface aefd?
D) What is the net flux through all five surfaces that enclose the shaded volume?

Answer 1

Concepts and reason

The concepts required to solve this problem are magnetic flux and the relation between magnetic field, surface area and the angle between the magnetic field lines and the normal to the area.

First, consider the expression for the magnetic flux across the surface ‘abcd’. Find the angle between the magnetic field and the normal. Substitute the values of magnetic field and area to calculate the magnetic flux.

Now, consider the surface ‘befc’. In this case, find the angle and then calculate the magnetic flux through the surface. Repeat the same method for the surface ‘aefd’ to find the magnetic flux. Finally the net flux can be calculated by adding the flux through each of the individual surfaces.

Fundamentals

Magnetic flux through the surface is defined as the “surface integral of the normal component of magnetic field passing through that surface”. It gives the number of magnetic field lines passing through a given area.

The expression for magnetic flux is given by,

$\Phi = BA\cos \theta$

Here, B is the magnetic field, A is the area, and $\theta$ is the angle between the magnetic field lines, and the normal to the area.

The magnetic flux across the surface ‘abcd’ is,

${\Phi _1} = B{A_1}\cos {\theta _1}$

From the given diagram in the problem, as shown surface ‘abcd’ direction of the magnetic field along z axis, since the angle between the magnetic field and normal to the surface is${\theta _1} = 90^\circ$.

Since area of the surface abcd is,

${A_1} = lb$

Here, ${A_1}$ is the area of the surface abcd, l is the length of the surface, and b is the breadth of the surface.

Convert the value of l from cm to m.

$\begin{array}{c}\\l = 30.0{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.0{\rm{ cm}}}}} \right)\\\\ = 0.3{\rm{ m}}\\\end{array}$

Convert the value of b from cm to m.

$\begin{array}{c}\\b = 40.0{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.0{\rm{ cm}}}}} \right)\\\\ = 0.4{\rm{ m}}\\\end{array}$

Substitute $0.128\,{\rm{T}}$ for B, $0.3{\rm{ m}}$ for l, $0.4{\rm{ m}}$ for b, and $90^\circ$ for $\theta$ to find ${\Phi _1}$.

$\begin{array}{c}\\{\Phi _1} = \left( {0.128\,{\rm{T}}} \right)\left[ {\left( {0.4{\rm{ m}}} \right)\left( {0.3{\rm{ m}}} \right)} \right]\left( 0 \right)\\\\ = 0\,{\rm{Wb}}\\\end{array}$

Magnetic flux across the surface befc is,

${\Phi _2} = B{A_2}\cos {\theta _2}$

Area of the square is,

${A_2} = {l^2}$

Hence the above expression becomes,

${\Phi _2} = B{l^2}\cos {\theta _2}$

Substitute $0.128\,{\rm{T}}$ for B, $0.3{\rm{ m}}$ for l, and $180^\circ$ for $\theta$ to find ${\Phi _2}$.

$\begin{array}{c}\\{\Phi _2} = \left( {0.128\,{\rm{T}}} \right){\left[ {0.3{\rm{ m}}} \right]^2}\left( {\cos 180^\circ } \right)\\\\ = - 0.01152\,{\rm{Wb}}\\\end{array}$

The angle between the magnetic field and the normal to the surface aefd is,

$\cos {\theta _3} = \frac{l}{h}$

Substitute l with 30 cm and h with 50 cm.

$\begin{array}{c}\\\cos {\theta _3} = \frac{{30{\rm{ cm}}}}{{50{\rm{ cm}}}}\\\\ = 0.6\\\end{array}$

Area of the surface aefd is,

${A_3} = lh$

Convert the value of h from cm to m.

$\begin{array}{c}\\h = 50{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)\\\\ = 0.5{\rm{ m}}\\\end{array}$

Substitute 0.3 m for l and 0.5 m for h to find ${A_3}$.

$\begin{array}{c}\\{A_3} = \left( {0.3\,{\rm{m}}} \right)\left( {0.5\,{\rm{m}}} \right)\\\\ = 0.15\,{{\rm{m}}^2}\\\end{array}$

Magnetic flux across the surface aefd is,

${\Phi _3} = B{A_3}\cos {\theta _3}$

Substitute $0.128\,{\rm{T}}$ for B, $0.15\,{{\rm{m}}^2}$ for ${A_3}$, and 0.6 for ${\theta _3}$ to find ${\Phi _3}$.

$\begin{array}{c}\\{\Phi _3} = \left( {0.128\,{\rm{T}}} \right)\left( {0.15\,{{\rm{m}}^2}} \right)\left( {0.6} \right)\\\\ = 0.01152\,{\rm{Wb}}\\\end{array}$

The net magnetic flux through all the five surfaces that encloses the shaded volume is,

${\Phi _{{\rm{total}}}} = {\Phi _1} + {\Phi _2} + {\Phi _3}$

Here, ${\Phi _1}$ is the magnetic flux across the surface ‘abcd’, ${\Phi _2}$ is the magnetic flux across the surface ‘befc’, and ${\Phi _3}$ is the magnetic flux across the surface ‘aefd’.

Substitute $0\,{\rm{Wb}}$ for ${\Phi _1}$, $- 0.01152\,{\rm{Wb}}$ for ${\Phi _2}$ and $0.01152\,{\rm{Wb}}$ for ${\Phi _3}$ to find ${\Phi _{{\rm{total}}}}$

$\begin{array}{c}\\{\Phi _{{\rm{total}}}} = \left( {0\,{\rm{Wb}}} \right) + \left( { - 0.01152\,{\rm{Wb}}} \right) + \left( {0.01152\,{\rm{Wb}}} \right)\\\\ = 0\,{\rm{Wb}}\\\end{array}$

Ans:

The magnetic flux across the surface ‘abcd’ is$0\,{\rm{Wb}}$.

The magnetic flux across the surface ‘befc’ is$- 0.01152\,{\rm{Wb}}$.

The magnetic flux across the surface ‘aefd’ is$0.01152{\rm{ Wb}}$.

The net magnetic flux through all the five surfaces that encloses the shaded volume is$0\,{\rm{Wb}}$.

Answer 2

The magnetic flux passing through the surface of an areaasφ=BAA.A(abcd)==0.4*0.3=0.12m²φ=0.128*1.2=0.01536Wbm²B.A(befc)=0.3*0.3=0.09m²C.A(aefd)=0.5*0.3=0.15m²D.A(abe)=1/2*0.3*0.4=0.06m²A(dcf)=1/2*0.3*0.4=0.06m²The total flux is given byφ=BA(abcd)+BA(befc)+BA(aefd)+BA(abe)+BA(dcf)Plug in the values to get the right solution.

Answer 3

This is exactly what I tried. B*A does not give me thecorrect answer for some reason. The only one that worked outwas B