First slide

Magnetic flux and induced emf

Question

A conducting rod PO of length l = 2 m is moving at a speed of 2 ms^-1 making an angle of 30^o with its length. A uniform magnetic field B = 2T exists in a direction perpendicular to the plane of motion. Then

Moderate

A

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

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

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

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

Solution

Emf induced across the rod AB is
$\begin{array}{l} e = B ℓ v \sin θ = 2 \times 2 \times 2 \times \sin 30 \\ \Rightarrow e = 4 V \end{array}$
Free electrons of the rod shift towards right due to force
$\vec{F} = q (\vec{v} \times \vec{B})$
Thus, end P is at higher potential
or V_P - V_Q= 4V.
Thus, choice (b) is correct.

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