A) \[{{V}_{p}}-{{V}_{Q}}=8V\]
B) \[{{V}_{p}}-{{V}_{Q}}=4V\]
C) \[{{V}_{Q}}-{{V}_{P}}=8V\]
D) \[{{V}_{Q}}-{{V}_{P}}=4V\]
Correct Answer: B
Solution :
[b] Emf induced across the rod AB is \[e=\vec{B}.(\vec{\ell }\times \vec{v})=B\ell v\sin \theta =2\times 2\times 2\times \sin 30\]\[e=4V\] Free electrons of the rod shift towards right due to force \[q(\vec{v}\times \vec{B})\] Thus, end P is at higher potential Or \[{{V}_{P}}-{{V}_{Q}}=4\,V.\] Thus, choice (2) is correct.
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