A horizontal wire free to slide on the vertical rails of a conducting frame as shown in figure. The wire has a mass m and length ℓ and the resistance of the circuit is R. If a uniform magnetic field B is directed perpendicular to the frame, then find the terminal speed of the wire as it falls under the force of gravity, in m/s. ( m =1 kg , R =1 Ω, B =1 T , ℓ=√2 m )

Question

A horizontal wire free to slide on the vertical rails of a conducting frame as shown in figure. The wire has a mass m and length ℓ and the resistance of the circuit is R. If a uniform magnetic field B is directed perpendicular to the frame, then find the terminal speed of the wire as it falls under the force of gravity, in m/s. ( m =1 kg , R =1 Ω, B =1 T , ℓ=√2 m ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d18be8c2623b2ef2ab22cccdbad9d8ab-.png" /> Power by ext. force = Rate of change of KE + electrical power As wire is moving with terminal velocity its KE = constant m g v T=i2 R mgv T =(( B ℓ v T / R ))2 × R mg =( B 2 ℓ2 v T / R ) v T =( mgR / B 2 ℓ2)

Power by ext. force = Rate of change of KE+ electrical power As wire is moving with terminal velocity its KE= constant mgvT=i2R mgvT​=(RBℓvT​​)2×R mg=RB2ℓ2vT​​ vT​=B2ℓ2mgR​

Power by ext. force $=$ Rate of change of $K E +$ electrical power
As wire is moving with terminal velocity its $K E =$ constant
$m gv T = i^{2} R$
$m g v_{T} = (\frac{B ℓ v _{T}}{R})^{2} \times R$
$m g = \frac{B ^{2} ℓ ^{2} v _{T}}{R}$
$v_{T} = \frac{m g R}{B ^{2} ℓ ^{2}}$