A conducting rod P Q of length l=2 m is moving at a speed of 2 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 P Q of length l=2 m is moving at a speed of 2 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 (A) VP-VQ=8 V (B) VP-VQ=4 V (C) VQ-VP=8 V (D) VQ-VP=4 V

Tardigrade Admin · Accepted Answer

Emf induced across the rod A B is e= vecB ⋅( vecℓ × vecv)=B ℓ v sin θ =2 × 2 × 2 × sin 30 e=4 V Free electrons of the rod shift towards right due to force q( vecv × vecB) Thus, end P is at higher potential or VP-VQ=4 V.

Q. A conducting rod $PQ$ of length $l = 2 m$ is moving at a speed of $2 m s^{- 1}$ making an angle of $3 0^{\circ}$ with its length. A uniform magnetic field $B = 2 T$ exists in a direction perpendicular to the plane of motion. Then

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

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

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

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

Emf induced across the rod $A B$ is
$e = B \cdot (ℓ \times v) = B ℓ v sin θ$
$= 2 \times 2 \times 2 \times sin 30$
$e = 4 V$
Free electrons of the rod shift towards right due to force
$q (v \times B)$
Thus, end $P$ is at higher potential or $V_{P} - V_{Q} = 4 V$ .

Q. A conducting rod PQ of length l=2m is moving at a speed of 2ms−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

VP​−VQ​=8V

VP​−VQ​=4V

VQ​−VP​=8V

VQ​−VP​=4V