SATHEE: Electro Magnetic Induction And Alternating Currents Question 407

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) $V_{p} - V_{Q} = 8 V$

B) $V_{p} - V_{Q} = 4 V$

C) $V_{Q} - V_{P} = 8 V$

D) $V_{Q} - V_{P} = 4 V$

Show Answer

Answer:

Correct Answer: B

Solution:

Emf induced across the rod AB is

$e = \vec{B} . (\vec{ℓ} \times \vec{v}) = B ℓ v \sin θ = 2 \times 2 \times 2 \times \sin 30$ $e = 4 V$

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.

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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,ms−1 making an angle of 30∘ with its length. A uniform magnetic field B = 2 T exists in a direction perpendicular to the plane of motion. Then

Options:

Answer:

Solution:

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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