question_answer

A Loop, made of straight edges has six corners at A(0, 0, 0), B(L, 0, 0) C(L, L, 0), D(0, L, 0). E(0, L, L) and F(0, 0, L). A magnetic field \[\overrightarrow{B}={{B}_{0}}(\widehat{i}+\widehat{k})T\] is present in the region. The flux passing through the loop ABCDEFA (in that order) is:                                  (NCERT EXEMPLAR)

A) \[{{B}_{0}}{{L}^{2}}Wb\]                 

B) \[2{{B}_{0}}{{L}^{2}}Wb\]

C) \[\sqrt{2}{{B}_{0}}{{L}^{2}}Wb\]                 

D) \[4{{B}_{0}}{{L}^{2}}Wb\]

Correct Answer: B

Solution :

(b) \[2{{B}_{0}}{{L}^{2}}Wb\] Here,     \[\overrightarrow{B}={{B}_{0}}(\widehat{i}+\widehat{k})T\] Area vector of \[ABCD={{L}^{2}}\widehat{k}\] Area vector \[DEFA={{L}^{2}}\widehat{i}\] Total area vector, \[\overrightarrow{A}={{L}^{2}}(\widehat{i}+\widehat{k})\] Total magnetic flux, \[\phi =\overrightarrow{B}\cdot \overrightarrow{A}\] \[={{B}_{0}}(\widehat{i}+\widehat{k})\cdot {{L}^{2}}(\widehat{i}+\widehat{k})={{B}_{0}}{{L}^{2}}(1+1)=2{{B}_{0}}{{L}^{2}}Wb\]