A conducting rod PQ of length is moving at a velocity making an angle with its length

Question

A conducting rod PQ of length $l = 2 m$ is moving at a velocity $v = 8 m s^{- 1}$ making an angle $30 °$ with its length. A uniform magnetic field $B = 3 T$ exists in a direction perpendicular to the plane of motion. Then,

Smart Achievers · Accepted Answer

(2)

$ξ = B l v_{⊥}$

$\Rightarrow ξ = (3) (2) (8 \sin 30)$

$\Rightarrow ξ = 24 V$

From Fleming’s Right Hand Rule, the induced current in the rod PQ is directed from Q to B, thus giving an equivalent emf replacement of the motional emf as

So, P is at a higher potential.

Solution

Q.
A conducting rod PQ of length $l = 2 m$ is moving at a velocity $v = 8 m s^{- 1}$ making an angle $30 °$ with its length. A uniform magnetic field $B = 3 T$ exists in a direction perpendicular to the plane of motion. Then,

1 $V_{P} - V_{Q} = 49 V$

2 $V_{P} - V_{Q} = 24 V$

3 $V_{Q} - V_{P} = 49 V$

4 $V_{Q} - V_{P} = 48 V$

Ans.
(2)

$ξ = B l v_{⊥}$

$\Rightarrow ξ = (3) (2) (8 \sin 30)$

$\Rightarrow ξ = 24 V$

From Fleming’s Right Hand Rule, the induced current in the rod PQ is directed from Q to B, thus giving an equivalent emf replacement of the motional emf as

So, P is at a higher potential.

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Solution

Q. A conducting rod PQ of length l=2m is moving at a velocity v=8ms-1 making an angle 30° with its length. A uniform magnetic field B=3T exists in a direction perpendicular to the plane of motion. Then,

1 VP-VQ=49V

2 VP-VQ=24V

3 VQ-VP=49V

4 VQ-VP=48V

Ans. (2) ξ=Blv⊥ ⇒ξ=(3)(2)(8sin30) ⇒ξ=24 V From Fleming’s Right Hand Rule, the induced current in the rod PQ is directed from Q to B, thus giving an equivalent emf replacement of the motional emf as So, P is at a higher potential.