A conducting rod AB of length l=1 m is moving at a velocity v=4 m s- 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

Question

A conducting rod AB of length l=1 m is moving at a velocity v=4 m s- 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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-3jb6dxr4v445.png" /> (A) vA-vB=8 v (B) vA-vB=4 v (C) vB-vA=8 v (D) vB-vA=4 v

Tardigrade Admin · Accepted Answer

e=B(v s i n (30)o)l=4 volt Fleming right hand Ii n d B to A ⇒ VA-VB=4 volt

Q. A conducting rod AB of length $l = 1 m$ is moving at a velocity $v = 4 m s^{- 1}$ 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

$v_{A} - v_{B} = 8 v$

$v_{A} - v_{B} = 4 v$

$v_{B} - v_{A} = 8 v$

$v_{B} - v_{A} = 4 v$

$e = B (v s in (30)^{o}) l = 4 v o lt$
Fleming right hand $I_{in d}$ B to A
$\Rightarrow V_{A} - V_{B} = 4 v o lt$

Q. A conducting rod AB of length l=1m is moving at a velocity v=4ms−1 making an angle of 30° with its length. A uniform magnetic field B=2T exists in a direction perpendicular to the plane of motion, then

vA​−vB​=8v

vA​−vB​=4v

vB​−vA​=8v

vB​−vA​=4v