In the figure shown PQRS is a fixed resistanceless conducting frame in a uniform and constant magnetic field of strength B. A rod EF of mass m length l and resistance R can smoothly move on this frame.A capacitor charged to a potential difference V0 initially is connected as shown in the figure.Find the velocity of the rod as function of time t if it is released at t=0 from rest.

Key Concepts You Must Know

1. Motional emf
   A conductor of length $l$ moving with speed $v$ perpendicular to a magnetic field $B$ develops an emf
   $\varepsilon = B l v$

2. Faraday–Lenz Rule → Direction of current
   The induced current always opposes the change that produces it (Lenz). Here, directions are chosen so that we do not worry about signs at the last minute; we only need consistency.

3. Newton’s 2nd law with magnetic force
   Current $I$ in a length $l$ placed in field $B$ feels force $F = I l B$ (right-hand rule). This force is what accelerates or decelerates the rod:
   $m \frac{dv}{dt} = - I l B$ (minus sign because the magnetic force opposes the motion that created the current).

4. Circuit equation with capacitor
   The loop contains three elements in series: motional emf ($\varepsilon$), rod resistance $R$, and capacitor voltage $q/C$. KVL gives $\varepsilon - I R - \frac{q}{C} = 0$ and current $I$ equals rate of change of capacitor charge: $I = dq/dt$.

5. Coupled differential equations
   You have one equation from mechanics and one from the circuit. Together they give a single first-order ODE for $v(t)$, which is solved just like a normal e^(–kt) decay problem.

6. Time constant
   The final expression will always look like $v(t) = v_{\infty}\bigl(1 - e^{-t/\tau}\bigr)$ where $v_{\infty}$ is the terminal speed and $\tau$ is the system’s time constant.

Logical chain of steps to solve

1. Write KVL around the loop → relate $v$, $q$, $I$.
2. Write Newton’s law → relate $v$, $I$.
3. Eliminate $I$ and $q$ to get a single ODE for $v(t)$.
4. Solve the ODE with the initial condition $v(0)=0$.
5. Express the final answer in tidy physics parameters ($B, l, R, C, V_0, m$).

What’s happening in simple words?

Imagine two long metal rails making a track and a metal rod that can slide on these rails like a train. The whole set-up is inside a steady magnetic field that points into the page.

1. Sliding rod = dynamo: When the rod moves, it cuts magnetic lines, so a battery-like push (emf) is produced along the rod.
2. A real capacitor is waiting: Before we let the rod go, we connected a capacitor that already has some charge (so it behaves like a small battery of its own).
3. Tug-of-war: The induced emf tries to send current one way; the capacitor’s voltage tries to send current the other way. The current that finally flows produces a magnetic force on the rod that can slow it down or speed it up.
4. Energy shuffle: Energy keeps shuffling between three stores —
   • Kinetic energy of the rod
   • Electric energy of the capacitor
   • Heat in the rod’s resistance

Because of that shuffle, the speed of the rod grows at first, then settles to a fixed value. The maths of that time-story is what we need to work out.

Step-by-step mathematical solution

Let

• $B$ = magnetic field strength (uniform, perpendicular to plane)
• $l$ = length of sliding rod EF
• $R$ = resistance of rod
• $m$ = mass of rod
• $C$ = capacitance of the capacitor
• $V_0$ = initial potential difference on capacitor
• $q(t)$ = charge on capacitor at time $t$ (positive on the plate connected as shown)
• $v(t)$ = speed of the rod at time $t$ (along rails)

Initial conditions: $v(0) = 0, \qquad q(0) = q_0 = C V_0$

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1. Circuit (KVL) equation

Induced emf $\varepsilon = B l v$

KVL around the loop gives

$B l v - I R - \frac{q}{C} = 0$

But current is the rate at which capacitor charge changes: $I = \frac{dq}{dt}$

So

B l v - R \frac{dq}{dt} - \frac{q}{C} = 0 \tag{1}

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2. Mechanical (Newton) equation

Magnetic force on the rod: $F = I l B$

Taking opposing direction as negative,

m \frac{dv}{dt} = - I l B = - B l \frac{dq}{dt} \tag{2}

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3. Eliminate $I$ and $q$

From (2)

$\frac{dv}{dt} = - \frac{B l}{m} \frac{dq}{dt} \implies dq = - \frac{m}{B l} dv$

Integrate using $v(0)=0$ and $q(0)=q_0$:

q = q_0 - \frac{m}{B l} v \tag{3}

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4. Substitute (3) in KVL (1)

First, find $dq/dt$ from (3): $\frac{dq}{dt} = - \frac{m}{B l} \frac{dv}{dt}$

Put $q$ and $dq/dt$ in (1):

B l v - R \left( - \frac{m}{B l} \frac{dv}{dt} \right) - \frac{1}{C}\left(q_0 - \frac{m}{B l} v \right) &= 0\\[4pt] \Rightarrow B l v + \frac{R m}{B l} \frac{dv}{dt} - \frac{q_0}{C} + \frac{m}{B l C} v &=0 \end{aligned}$$ Group like terms: $$\frac{R m}{B l} \frac{dv}{dt} + \left(B l + \frac{m}{B l C}\right) v - \frac{q_0}{C} = 0$$ Define $$\alpha = \frac{(B l)^2 + \frac{m}{C}}{R m}, \qquad \beta = \frac{B l V_0}{R m}$$ (using $q_0 = C V_0$). Equation becomes $$\frac{dv}{dt} + \alpha v = \beta$$ --- #### 5. Solve first-order ODE Standard integrating-factor solution: $$v(t) = \left[ v(0) - \frac{\beta}{\alpha} \right] e^{-\alpha t} + \frac{\beta}{\alpha}$$ Since $v(0)=0$: $$v(t) = \frac{\beta}{\alpha}\bigl(1 - e^{-\alpha t}\bigr)$$ Substitute $\alpha$ and $\beta$: $$\boxed{\displaystyle v(t) = \frac{B l V_0}{(B l)^2 + \dfrac{m}{C}} \left( 1 - e^{-\left( \dfrac{(B l)^2 + \dfrac{m}{C}}{R m} \right) t} \right) }$$ --- #### 6. Physical interpretation * **Terminal speed** (as $t \to \infty$): $$v_{\infty} = \frac{B l V_0}{(B l)^2 + m/C}$$ * **Time constant:** $$\tau = \frac{R m}{(B l)^2 + m/C}$$ Faster decay if the magnetic field or capacitance is large, or if the rod is light.