Let's tackle the problem step by step for both cases: **Given:** - Conductive rod of mass \( m \), resistance \( R \), on frictionless, resistance-free rails spaced \( l \) apart. - Uniform magnetic field \( B \), perpendicular to the rod and rails. - At \( t = 0 \), the rod is at rest. A voltage source is connected at points \( a \) and \( b \). **To Find:** - Velocity \( v(t) \) of the rod as a function of time for: - (a) Constant current \( I \) - (b) Constant voltage \( U_0 \) - (c) For each case, does the rod reach a steady-state velocity? What is it? --- ## (a) **Constant Current \( I \)** ### **Analysis** - The current \( I \) in the rod creates a Lorentz force: \[ F = I l B \] (Direction by right-hand rule: if current flows along the rod, force is along rails.) - Newton's 2nd law: \[ m \frac{dv}{dt} = I l B \] This is a **constant force**. ### **Solution** \[ \frac{dv}{dt} = \frac{I l B}{m} \] Integrate both sides: \[ v(t) = \int_0^t \frac{I l B}{m} dt = \frac{I l B}{m} t \] **Initial condition:** \( v(0) = 0 \) **Final answer (a):** \[ \boxed{v(t) = \frac{I l B}{m} t} \] ### **Steady-State?** - The velocity increases *linearly* with time; *no steady-state* is reached (ignoring relativistic effects and infinite rails). --- ## (b) **Constant Voltage \( U_0 \)** ### **Analysis** - The voltage across the rod is \( U_0 \). - As the rod moves, it induces a motional emf: \[ \varepsilon_{mot} = B l v \] - The total voltage across the rod and source: \[ U_0 = IR + \varepsilon_{mot} \] So, \[ I = \frac{U_0 - B l v}{R} \] - The force on the rod: \[ F = I l B = \left( \frac{U_0 - B l v}{R} \right) l B \] \[ F = \frac{l B}{R} (U_0 - B l v) \] - Newton's 2nd law: \[ m \frac{dv}{dt} = \frac{l B}{R} (U_0 - B l v) \] \[ \frac{dv}{dt} + \frac{(B l)^2}{m R} v = \frac{l B U_0}{m R} \] This is a **first-order linear ODE**. ### **Solution** Let - \( \alpha = \frac{(B l)^2}{m R} \) - \( \beta = \frac{l B U_0}{m R} \) The ODE: \[ \frac{dv}{dt} + \alpha v = \beta \] The solution (with \( v(0) = 0 \)): \[ v(t) = \frac{\beta}{\alpha} (1 - e^{-\alpha t}) \] Plug in \( \beta/\alpha \): \[ \frac{\beta}{\alpha} = \frac{\frac{l B U_0}{m R}}{\frac{(B l)^2}{m R}} = \frac{U_0}{B l} \] **Final answer (b):** \[ \boxed{ v(t) = \frac{U_0}{B l} \left(1 - e^{-\frac{(B l)^2}{m R} t}\right) } \] ### **Steady-State?** - As \( t \to \infty \), \( e^{-\alpha t} \to 0 \), - \[ v_{\text{final}} = \frac{U_0}{B l} \] --- ## (c) **Steady-State Velocity?** - **(a) Constant current:** No steady-state velocity; \( v(t) \) increases without bound. - **(b) Constant voltage:** Yes, steady-state velocity: \[ v_{\text{final}} = \frac{U_0}{B l} \] --- ## **Summary Table** | Case | \( v(t) \) | Steady-State? | \( v_{\text{final}} \) | |----------------------------|-------------------------------------------------------------------|:-------------:|-----------------------------| | (a) Constant current \( I \) | \( \displaystyle v(t) = \frac{I l B}{m} t \) | No | None | | (b) Constant voltage \( U_0 \) | \( \displaystyle v(t) = \frac{U_0}{B l} \left[1 - e^{-\frac{(B l)^2}{m R} t}\right] \) | Yes | \( \frac{U_0}{B l} \) | --- ### **Physical Interpretation** - For constant current, the rod is constantly accelerated (unphysical for long times, but possible in idealized case). - For constant voltage, the rod asymptotically approaches a terminal velocity where the motional emf cancels the applied voltage.