Two point masses, m each carrying charges - q and + q are attached to the ends of a massless rigid non-conducting wire of length L . When this arrangement is placed in a uniform electric field, then it deflects through an angle θ . The minimum time needed by the rod to align itself along the field is

Question

Two point masses, m each carrying charges - q and + q are attached to the ends of a massless rigid non-conducting wire of length L . When this arrangement is placed in a uniform electric field, then it deflects through an angle θ . The minimum time needed by the rod to align itself along the field is, 2 π m L q E, π 2 m L 2 q E, π 2 m L q E, 2 π 3 m L q E

MARKS App · Accepted Answer

When the wire is brought in a uniform field E, then the torque is given by τ = q E L sin ⁡ θ = q E L θ ∵ θ i s v e r y s m a l l The moment of inertia of rod AB about O is I = m L 2 2 + m L 2 2 = m L 2 2 As τ = I α So, α = τ I = 2q E L θ m L 2 ⇒ ω 2 θ = 2 q E L θ m L 2 ∵ α = ω 2 θ ⇒ ω 2 = 2 q E m L The time period of the wire is T = 2 π ω = 2 π m L 2 2 q E The rod will become parallel to the field in time T 4 So, t = T 4 = π 2 m L 2 q E

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