Electro Magnetic Induction And Alternating Currents Question 311
Question: At a place the value of horizontal component of the earth’s magnetic field H is $ 3\times {{10}^{-5}},Weber/m^{2} $ . A metallic rod AB of length 2 m placed in east-west direction, having the end A towards east, falls vertically downward with a constant velocity of 50 m/s. Which end of the rod becomes positively charged and what is the value of induced potential difference between the two ends [MP PET 1996]
Options:
A) End A, $ 3\times {{10}^{-3}},mV $
B) End A, 3 mV
C) End B, $ 3\times {{10}^{-3}},mV $
D) End B, 3 mV
Show Answer
Answer:
Correct Answer: B
Solution:
Induced potential difference between two ends $ =Blv=B_{H}lv $ $ =3\times {{10}^{-5}}\times 2\times 50=30\times {{10}^{-3}}volt=3,millivolt $ By Fleming’s right hand rule, end A becomes positively charged.
BETA
