Question

magnetic flux through a square of side 6cm placed with it's plane making angle 60° with a uniform field of 5T will be​

Answer 1

To find:

Magnetic flux through a square of side 6cm placed with it's plane making angle 60° with a uniform field of 5 Tesla.

Calculation:

General expression for magnetic flux through a specified area is given as:

$\boxed{ \sf{ \phi = B \times A \times \cos( \theta) }}$

Note that $\theta$ is the angle between the magnetic field intensity and the area vector.

So, $\theta$ in this case is :

$\therefore \: \theta = {90}^{ \circ} - {60}^{ \circ} = {30}^{ \circ}$

Now , putting all the available values in SI unit:

$\sf{ \phi = B \times A \times \cos( \theta) }$

$= > \sf{ \phi = 5 \times { (\dfrac{6}{100}) }^{2} \times \cos( {30}^{ \circ} ) }$

$= > \sf{ \phi = 5 \times { (\dfrac{6}{100}) }^{2} \times \dfrac{ \sqrt{3} }{2} }$

$= > \sf{ \phi = 5 \times \dfrac{36}{ {10}^{4} } \times \dfrac{ \sqrt{3} }{2} }$

$= > \sf{ \phi = 5 \times \dfrac{18}{ {10}^{4} } \times \sqrt{3} }$

$= > \sf{ \phi = \dfrac{90}{ {10}^{4} } \times \sqrt{3} }$

$= > \sf{ \phi = \dfrac{9}{ {10}^{3} } \times \sqrt{3} }$

$= > \sf{ \phi = 9 \sqrt{3} \times {10}^{ - 3} \:T {m}^{2} }$

So, final answer is:

$\boxed{ \red{ \large{ \sf{ \phi = 9 \sqrt{3} \times {10}^{ - 3} \:T {m}^{2} }}}}$

Answer 2

$\sf \mathfrak{\underline{\red{Given:-}}}$

• A square of side 6cm is placed with it's plane making angle 60° with a uniform field of 5T.

$\sf \mathfrak{\underline{\blue{To\: find:-}}}$

• Magnetic flux through it.

$\sf \mathfrak{\underline{\green{Knowledge\: required:-}}}$

• The magnetic flux through a surface is the component of the magnetic field passing through that surface.

↝ General expression for magnetic flux through a specified area is ,

$✷\underline{\boxed{\pink{\sf{ \phi = B \times A \times \cos( \theta) }}}}</p><p>$

Where,

↗ B is the magnitude of the magnetic field (having the unit of Tesla, T).

↗A is the area of the surface.

↗ θ is the angle between the magnetic field lines and the normal (perpendicular) to A.

$\sf \mathfrak{\underline{\red{Solution:-}}}$

Value of 'θ' will be :

$\implies \theta = {90}^{ \circ} - {60}^{ \circ} = {30}^{ \circ }$

Now ,

➝ ϕ=B×A×cos(θ)

$➝ \sf{ \phi = 5 \times { (\dfrac{6}{100}) }^{2} \times \cos( {30}^{ \circ} ) }\\\\\\</p><p></p><p>➝ \sf{ \phi = 5 \times { (\dfrac{6}{100}) }^{2} \times \dfrac{ \sqrt{3} }{2} }\\\\\\</p><p></p><p> ➝ \sf{ \phi = 5 \times \dfrac{36}{ {10}^{4} } \times \dfrac{ \sqrt{3} }{2} }\\\\\\</p><p>➝ \sf{ \phi = 5 \times \dfrac{18}{ {10}^{4} } \times \sqrt{3} }\\\\\\</p><p>➝ \sf{ \phi = \dfrac{90}{ {10}^{4} } \times \sqrt{3} }\\\\\\ ➝\sf{ \phi = \dfrac{9}{ {10}^{3} } \times \sqrt{3} }\\\\\\</p><p></p><p>➜ \sf{ \phi = 9 \sqrt{3} \times {10}^{ - 3} \:T {m}^{2} }</p><p>[tex]$

Therefore,

$\underline{\boxed{ \pink{ \large{ \sf{ \implies \phi = 9 \sqrt{3} \times {10}^{ - 3} \:T {m}^{2} }}}}}$

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