A) \[{{v}_{A}}-{{v}_{E}}=\frac{9}{2}B\omega \,{{i}^{2}}\]
B) \[{{v}_{E}}-{{v}_{A}}=\frac{9}{2}B\omega \,{{i}^{2}}\]
C) \[{{v}_{D}}-{{v}_{E}}=\frac{B\omega \,{{i}^{2}}}{2}\]
D) \[{{v}_{A}}-{{v}_{E}}=4B\omega \,{{i}^{2}}\]
Correct Answer: D
Solution :
Since, \[{{V}_{E}}-{{v}_{D}}=\frac{B\omega l}{2}\times l\] \[{{v}_{E}}-{{v}_{D}}=\frac{B\omega {{l}^{2}}}{2}\] ?. (i) \[{{v}_{E}}-{{v}_{A}}=\frac{B\omega 3l}{2}\times 3l=\frac{9B\omega {{l}^{2}}}{2}\] .... (ii) Subtracting Eq.(i) from Eq. (ii), we get. \[{{v}_{A}}-{{v}_{E}}=4B\omega {{l}^{2}}\] \[\Rightarrow \] \[{{v}_{A}}>{{v}_{E}}\]You need to login to perform this action.
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