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  •  In figure there is a ring of area <img alt="A = 10 m^2" sr

 In figure there is a ring of area A = 10 m^2  is kept in a constant magnetic field  B of intensity 1 Tesla 

What will be net flux through the surface 

Option: 1

0


Option: 2

10


Option: 3

20


Option: 4

15


Answers (1)

best_answer

As we have learned

Net flux through the surface -

\phi = \oint \vec{B}\cdot \vec{dA}= BA\cos \Theta
 

- wherein

\phi = Magnetic Flux

B = Magnetic field 

\Theta = Angle between area vector and magnetic field vector

 

 Magnetic  flux through any surface is , 

\phi = \oint \vec B \cdot d \vec A= BA \cos \theta

 

From figure: the angle b/w magnetic field and an ring element d\vec A \: is \: 90 \degree

So \phi = 0

as \cos 90 \degree = 0

 

 

 

 

 

 

 

Posted by

rishi.raj

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