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, \[\operatorname{Q}=B.A={{B}_{0}}(\hat{i}+\hat{k}).{{L}^{2}}(\hat{i}+\hat{k})={{B}_{0}}{{\operatorname{L}}_{2}}\] |
\[[\hat{i}.\hat{i}+\hat{i}.\hat{k}+\hat{k}.\hat{i}+\hat{k}.\hat{k}]\] |
=\[{{B}_{0}}{{\operatorname{L}}_{2}}[1+0+0+1]\] \[(\therefore \cos 9{{0}^{o}}=0)\] |
=\[2{{B}_{0}}{{\text{L}}^{2}}\text{Wb}\] |
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