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  • 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)

 

Answer: The answer is the option (b)

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

 

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