Electro Magnetic Induction And Alternating Currents Question 407
Question: A conducting rod PQ of length $ l $ = 2 m is moving at a speed of $ 2,m{{s}^{-1}} $ making an angle of $ 30{}^\circ $ with its length. A uniform magnetic field B = 2 T exists in a direction perpendicular to the plane of motion. Then
Options:
A) $ V_{p}-V_{Q}=8V $
B) $ V_{p}-V_{Q}=4V $
C) $ V_{Q}-V_{P}=8V $
D) $ V_{Q}-V_{P}=4V $
Show Answer
Answer:
Correct Answer: B
Solution:
-
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.
