A conducting rod PQ of length l-2 m is moving at a speed of 2 ms-1 making an angle of 60- with its length- A uniform magnetic field B-3 T exists in a direction perpendicular to the plane of motion- Then

Potential difference between points P and Q , V_P V_Q=e From the diagram, nbsp; nbsp;e=B (lsinθ) v After rotation, e=B (l sin 60^∘) v ⇒ e=√(3)× 2×√(3)2 ...

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Standard XII

Physics

Magnetic Field Due to a Straight Long Wire

A conducting ...

Question

A conducting rod $PQ$ of length $l = 2 m$ is moving at a speed of $2 {ms}^{- 1}$ making an angle of $60^{\circ}$ with its length. A uniform magnetic field $B = \sqrt{3} T$ exists in a direction perpendicular to the plane of motion. Then

V_{P} - V_{Q} = 4 V

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V_{P} - V_{Q} = 6 V

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V_{P} - V_{Q} = 8 V

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V_{P} - V_{Q} = 10 V

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Solution

The correct option is B $V_{P} - V_{Q} = 6 V$

Potential difference between points $P$ and $Q$ , $V_{P} - V_{Q} = e$

From the diagram, $e = B (l sin θ) v$

After rotation, $e = B (l sin 60^{\circ}) v$

$\Rightarrow e = \sqrt{3} \times 2 \times \frac{\sqrt{3}}{2} \times 2$

$\Rightarrow e = 6 V$

Hence, option (b) is the correct answer.

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A conducting rod PQ of length l=2 m is moving at a speed of 2 ms−1 making an angle of 60∘ with its length. A uniform magnetic field B=√3 T exists in a direction perpendicular to the plane of motion. Then

The correct option is B VP−VQ=6 V Potential difference between points P and Q , VP−VQ=e From the diagram, e=B (lsinθ) v After rotation, e=B (l sin 60∘) v ⇒e=√3×2×√32×2 ⇒e=6 V Hence, option (b) is the correct answer.

A conducting rod $PQ$ of length $l = 2 m$ is moving at a speed of $2 {ms}^{- 1}$ making an angle of $60^{\circ}$ with its length. A uniform magnetic field $B = \sqrt{3} T$ exists in a direction perpendicular to the plane of motion. Then