A) \[a/\sqrt{3}\]
B) 4
C) 1
D) \[a\sqrt{3}\]
Correct Answer: C
Solution :
Current density \[k\ge 1/2\] From Ampere's circuital law \[-1/2\le k\le 1/2\] For \[k\le 1/2\] \[2x+3y+z=1\] \[x+3y+2z=2\] \[cos\text{ }\alpha \] At \[1/\sqrt{3}\] \[1/\sqrt{2}\] For\[{{x}^{2}}={{y}^{2}}+xy\frac{dy}{dx}\]\[{{x}^{2}}={{y}^{2}}+3xy\frac{dy}{dx}\]\[{{y}^{2}}={{x}^{2}}+2xy\frac{dy}{dx}\]\[{{y}^{2}}={{x}^{2}}-2xy\frac{dy}{dx}\] At\[{{p}^{2}}+{{q}^{2}}=1,\] \[(p+q)\] So, \[\frac{1}{2}\]You need to login to perform this action.
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