SATHEE: Electro Magnetic Induction And Alternating Currents Question 311

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 $3 \times 10^{- 5}, W e b e r / m^{2}$ . 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, $3 \times 10^{- 3}, m V$

B) End A, 3 mV

C) End B, $3 \times 10^{- 3}, m V$

D) End B, 3 mV

Show Answer

Answer:

Correct Answer: B

Solution:

Induced potential difference between two ends $= B l v = B_{H} l v$ $= 3 \times 10^{- 5} \times 2 \times 50 = 30 \times 10^{- 3} v o l t = 3, m i l l i v o l t$ By Fleming’s right hand rule, end A becomes positively charged.

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Electro Magnetic Induction And Alternating Currents Question 311

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