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.



NCERT Chapter Video Solution

Dual Pane