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 B= \[{{B}_{0}}\text{ }(\hat{i}+\hat{k})\] Testa is present in the region. The flux passing through the loop ABCDEFA (in that order) is

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 :

Option [b] is correct.

Explanation: The loop can be considered in two planes:

(i) Plane of ABCDA is in X-Y plane. So its vector

\[\vec{A}\]is in Z-direction. Hence,

\[{{\operatorname{A}}_{1}}=|A|\hat{k}={{L}^{2}}\hat{k}\]

(ii) Plane of DEFAD is in Y-Z plane

So \[{{\operatorname{A}}_{2}}=|A|\hat{i}={{L}^{2}}\hat{i}\]

\[\therefore \]   \[A={{\operatorname{A}}_{1}}+{{A}_{2}}={{L}^{2}}(\hat{i}+\hat{k})\]

\[\operatorname{B}={{B}_{0}}(\hat{i}+\hat{k})\]

So, \[\operatorname{Q}=B.A={{B}_{0}}(\hat{i}+\hat{k}).{{L}^{2}}(\hat{i}+\hat{k})={{B}_{0}}{{\operatorname{L}}_{2}}\]

\[[\hat{i}.\hat{i}+\hat{i}.\hat{k}+\hat{k}.\hat{i}+\hat{k}.\hat{k}]\]

=\[{{B}_{0}}{{\operatorname{L}}_{2}}[1+0+0+1]\]         \[(\therefore \cos 9{{0}^{o}}=0)\]

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