SATHEE: Electro Magnetic Induction And Alternating Currents Question 410

Electro Magnetic Induction And Alternating Currents Question 410

Question: A metal conductor of length 1 m rotates vertically about one of its ends at an angular velocity 5 radians per second. If the horizontal component of the earth’s magnetic field is $0.2 \times 10^{- 4} T$ , then the emf developed between the two ends of the conductor is

Options:

A) $5, μ V$

B) $50, μ V$

C) $5, m V$

D) $50, m V$

Show Answer

Answer:

Correct Answer: B

Solution:

$E = \frac{1}{2} B ω R^{2} = \frac{1}{2} \times 0.2 \times 10^{- 4} \times 5 \times 1 \times 1$

$= 0.5 \times 10^{- 4} V = 54 \times 10^{- 1} V$

$= 50 \times 10^{- 6} V = 50 μ V$

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Electro Magnetic Induction And Alternating Currents Question 410

Question: A metal conductor of length 1 m rotates vertically about one of its ends at an angular velocity 5 radians per second. If the horizontal component of the earth’s magnetic field is 0.2×10−4T , then the emf developed between the two ends of the conductor is

Options:

Answer:

Solution:

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Question: A metal conductor of length 1 m rotates vertically about one of its ends at an angular velocity 5 radians per second. If the horizontal component of the earth’s magnetic field is $0.2 \times 10^{- 4} T$ , then the emf developed between the two ends of the conductor is