Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A loop, made of straight edges has six corners at A(0,0,0), B(L,O,0) C(L,L,0), D(0,L,0) E(0,L,L) and F(0,0,L). A magnetic field B = Bo (i + k) T is present in the region. The flux passing through the loop ABCDEFA (in that order) is

A loop, made of straight edges has six corners at A(0,0,0), B(L,O,0) C(L,L,0), D(0,L,0) E(0,L,L) \: and\: F(0,0,L). A magnetic field B = B_o (i + k) Tis present in the region. The flux passing through the loop ABCDEFA (in that order) is

(a) B_o\: L^2 Wb.

(b) 2 B_o L^2 Wb.

(c) \sqrt{}2 B_o L^2 Wb.

(d) 4 \: B_o L^2 Wb.

Answers (1)

The correct answer is the option b

Explanation:-

In this problem first, we have to analyze the area vector, loop ABCDA lies in an x-y plane whose area vector \vec{A_1}= L^2\hat{k} whereas loop ADEFA lies in the y-z plane whose area vector \vec{A_2}= L^2\hat{i}

And magnetic flux is  :  

\\ \phi _m=\vec{B}.\vec{A}\\ \vec{A}=\vec{A_1}+\vec{A_2}=\left ( L^2\hat{k} +L^2\hat{i}\right )\\ and\: \vec{B}=B_0\left ( \hat{i}+\hat{k} \right )\\ Now,\phi _m=\vec{B}.\vec{A}= B_0\left ( \hat{i}+\hat{k}\right ).\left ( L^2\hat{k} +L^2\hat{i}\right )\\ =2B_0L^2Wb

 

Posted by

infoexpert21

View full answer

Similar Questions

Latest Question