Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by B = B_o (2i + 3j + 4k)T where Bo is constant. The magnitude of flux passing through the square is

A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by B = B_o (2i + 3j + 4k)T where Bo is constant. The magnitude of flux passing through the square is

(a) 2 B_o L^2 Wb.
(b) 3 B_o L^2 Wb.
(c) 4 B_o L^2 Wb.
(d) \sqrt{}2 9B_o L^2 Wb.

 

Answers (2)

Answer: The answer is the option (c)

For elementary area dA of a surface, flux linked d\phi=BdA\cos \theta\:or\: d\phi=\vec{B}.\vec{dA}

So-net flux through the surface

\phi = \oint \vec{B}\times \vec{dA}=BA \cos \theta

In this problem
A= L^2\hat{k}\:and\:B=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )T\\ \phi = \vec{B}.\vec{A}=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right ).L^2\hat{k}= 4 B_o L^2 Wb.

Posted by

infoexpert21

View full answer

Answer: The answer is the option (c)

For elementary area dA of a surface, flux linked d\phi=BdA\cos \theta\:or\: d\phi=\vec{B}.\vec{dA}

So-net flux through the surface

\phi = \oint \vec{B}\times \vec{dA}=BA \cos \theta

In this problem

A= L^2\hat{k}\:and\:B=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )T\\ \phi = \vec{B}.\vec{A}=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right ).L^2\hat{k}= 4 B_o L^2 Wb.

Posted by

Neha Barik

View full answer

Similar Questions

Latest Question