Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACMotional & Rotational EMFMedium3 minQB

PhysicsMediummultiple choice

A bent rod PQR ( $PQ = QR = ℓ)$ shown here is rotating about its end P with angular speed $ω$ in a region of transverse magnetic field of strength B.

diagram

Answer:

B, D

Solution:

The bent rod is equivalent to the straight rod PR rotating the same way; with the $6 0^{\circ}$ bend, $PR = ℓ$ .

$V_{R} - V_{P} = \frac{1}{2} B ω ℓ^{2}$ - (B) correct.

Q is also at distance $ℓ$ from P: $V_{Q} - V_{P} = \frac{1}{2} B ω ℓ^{2} \Rightarrow V_{Q} - V_{R} = 0$ - (D) correct.

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Motional & Rotational EMF

⏱ 3mℹ️ Source: QB

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