Let's solve part (d) step-by-step with explanations: --- ### **Given:** - Mass of bar = \( m \) - Distance between rails = \( l \) - Magnetic field \( B \) (into the page) - Resistor \( R \) connected across rails - Bar moves to the right with speed \( v \) --- ### **(i) What is the current in the resistor? In what direction does it flow?** #### **Step 1: Find the induced emf** When the bar moves with speed \( v \) to the right in a magnetic field \( B \) (into the page), the induced emf (\( \mathcal{E} \)) across the bar is: \[ \mathcal{E} = B l v \] #### **Step 2: Find the current using Ohm's law** \[ I = \frac{\mathcal{E}}{R} = \frac{B l v}{R} \] #### **Step 3: Direction of current (using Fleming’s right-hand rule)** - Magnetic field: Into the page (\( \odot \)) - Velocity: To the right - Induced current: Up the bar (on the bar), so in the resistor it flows **from bottom rail to top rail** (counterclockwise in the loop). --- ### **(ii) What is the magnetic force on the bar? In what direction?** #### **Step 1: Force on a current-carrying conductor** \[ F = I l B \] Substitute \( I \) from above: \[ F = \left(\frac{B l v}{R}\right) l B = \frac{B^2 l^2 v}{R} \] #### **Step 2: Direction of force** Use Fleming’s left-hand rule: - Current: Up the bar - Magnetic field: Into the page => Force is to the **left** (opposite to the velocity) --- ### **(iii) If the bar starts with speed \( v_0 \) at \( t=0 \) and is left to slide, what is its speed at time \( t \)?** #### **Step 1: Equation of motion using Newton’s 2nd law** \[ m \frac{dv}{dt} = -F = -\frac{B^2 l^2 v}{R} \] \[ \frac{dv}{dt} = -\frac{B^2 l^2}{mR} v \] #### **Step 2: Solve differential equation** This is a separable ODE: \[ \frac{dv}{v} = -\frac{B^2 l^2}{mR} dt \] Integrate both sides: \[ \int_{v_0}^{v} \frac{dv}{v} = -\frac{B^2 l^2}{mR} \int_{0}^{t} dt \] \[ \ln\left(\frac{v}{v_0}\right) = -\frac{B^2 l^2}{mR} t \] \[ v = v_0 \exp\left(-\frac{B^2 l^2}{mR} t\right) \] --- ### **(iv) Check that the energy delivered to the resistor is exactly \(\frac{1}{2}mv_0^2\)** #### **Step 1: Energy delivered to the resistor** Energy delivered (\( E \)) is the total Joule heating: \[ E = \int_0^{\infty} I^2 R \, dt \] Recall, \( I = \frac{B l v}{R} \): \[ E = \int_0^{\infty} \left(\frac{B l v}{R}\right)^2 R \, dt = \int_0^{\infty} \frac{B^2 l^2 v^2}{R} dt \] Now, substitute \( v = v_0 \exp\left(-\frac{B^2 l^2}{mR} t\right) \): \[ E = \frac{B^2 l^2}{R} \int_0^{\infty} v_0^2 \exp\left(-2\frac{B^2 l^2}{mR} t\right) dt \] \[ E = \frac{B^2 l^2 v_0^2}{R} \int_0^{\infty} \exp\left(-2\frac{B^2 l^2}{mR} t\right) dt \] Let \( a = 2\frac{B^2 l^2}{mR} \) \[ E = \frac{B^2 l^2 v_0^2}{R} \cdot \frac{1}{a} = \frac{B^2 l^2 v_0^2}{R} \cdot \frac{mR}{2B^2 l^2} = \frac{m v_0^2}{2} \] #### **Step 2: Conclusion** The energy delivered to the resistor is **exactly the initial kinetic energy of the bar**. --- ## **Summary of Answers:** 1. **Current in resistor:** \( I = \frac{B l v}{R} \), flows counterclockwise (up the bar, through resistor from bottom to top rail). 2. **Magnetic force on bar:** \( F = \frac{B^2 l^2 v}{R} \), acts to the **left** (opposite to velocity). 3. **Speed at time \( t \):** \( v = v_0 \exp\left(-\frac{B^2 l^2}{mR} t\right) \) 4. **Energy delivered to resistor:** \( \frac{1}{2} m v_0^2 \) (matches initial kinetic energy).