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A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is \vec B=(3\hat{i}+4\hat{j})T.The quantity of flux through the loop ABCDEFA (in Wb) is:-    
Option: 1 175
Option: 2 60
Option: 3 25
Option: 4 170
 

Answers (1)

best_answer

 

 

Magnetic flux - - wherein

Magnetic flux-

The total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to
the magnetic flux linked with that area.

I.e for the below figure

Net magnetic flux through the surface is given by

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

where 

\phi_B= Magnetic Flux

B = Magnetic field 

\Theta = The angle between area vector and magnetic field vector

 

 

  • Magnetic flux is a scalar quantity.
  • Unit of magnetic flux -

          It's S.I. unit is Weber (wb) or Tesla\times m^2 and its C.G.S. unit is maxwell(Mx).

      and 1 \ w b=1 \ T m^{2}  and 1 \ Mx = 10^{-8 } \ wb

  • The dimension of magnetic flux is ML^{2}T^{-2}A^{-1} 
  • if θ = 0 then \phi = BA and  Flux will be positive.
  • If  \theta =\frac{\pi }{2} then Flux will be zero (i.e \phi = 0)

            

 

 

 

 

 

 

\vec B=(3\hat{i}+4\hat{j})T

Total area vectot=area of ABCD+area of DEFA=5^2\hat{k}+5^2\hat{i}=25 (\hat{i}+\hat{k})

Total Magnetic flux=\vec{B}.\vec{A}=(3\hat{i}+4\hat{j}).(25 (\hat{i}+\hat{k}))=(75+100)wb=175 wb

 

So option 3 will be correct answer.

Posted by

Ritika Jonwal

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