Physics · Viscosity, Stoke's law, terminal velocity, streamline and turbulent flow, critical velocity
Q Type your question. negligible resistance (Fig 3.126). The rails are connected
Q Type your question. negligible resistance (Fig 3.126). The rails are connected to each other at the bottom by a resistanceless rail paralle to the wire so that the wire and the rails form a closed rectangular conducting loop. The plane of the rails makes an angle \( \theta \) with the horizontal and a uniform vertical magnetic field of induction B exist throughout the region. Find the steady-state velocity of the wire.
- A. \( m g=\sin \theta \)
- B. \( \frac{m g}{R} \frac{\sin ^{2} \theta}{B^{2} l^{2} \cos ^{2} \theta} \)
- C. \( \frac{m g R \sin \theta}{B^{2} l^{2} \cos ^{2} \theta} \)
- D. \( \operatorname{mgr} \frac{\sin ^{2} \theta}{B^{2} / 2 \cos \theta} \)
Step-by-step solution
The induced EMF in the wire as it slides down the inclined rails is \(\mathcal{E} = B l v \cos \theta\). The current through the wire (resistance \(R\)) is \(I = \mathcal{E}/R = (B l v \cos \theta)/R\). The magnetic force on the wire is \(F_m = I l B = (B^2 l^2 v \cos^2 \theta)/R\), directed opposite to the motion (up the incline). The gravitational force component along the incline is \(mg \sin \theta\) downward. At steady state, net force is zero: \(mg \sin \theta = (B^2 l^2 v \cos^2 \theta)/R\). Solving for \(v\) gives \(v = \frac{mgR \sin \theta}{B^2 l^2 \cos^2 \theta}\).
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