A metal bar AB can slide on two parallel thick metallic rails separated by a distance l. A resistanc

Q: Why is the induced EMF equal to $d\phi/dt$ and not $-d\phi/dt$?

The negative sign in Faraday's law indicates the direction of the induced EMF (Lenz's Law). When applying Kirchhoff's Voltage Law, we assign the polarity of the induced EMF such that it drives the current $i$, so we use its magnitude $d\phi/dt$ as the voltage source.

Q: How did we get the expression for magnetic field $B$?

The magnetic field produced by a long straight wire carrying current $I_0$ at a distance $x$ is given by Ampere's Law as $B = \frac{\mu_0 I_0}{2\pi x}$.

Q: Why does the integral of $i dt$ give $\Delta q$?

By definition, electric current $i$ is the rate of flow of charge, $i = dq/dt$. Therefore, integrating current over a time interval gives the total charge $\Delta q$ that has passed through the circuit.

Q: Why is the initial current $i(0)$ zero?

The circuit contains an inductor $L$, which opposes any sudden change in current. Since the current was zero before the bar started moving, it must remain zero exactly at $t=0$.

Q: What happens to the circuit when the bar suddenly stops?

When the bar stops, its velocity becomes zero, so the motional EMF ($Bvl$) instantly becomes zero. The circuit then acts as a closed loop with only a resistor and an inductor, causing the existing current to decay exponentially.

Question

A metal bar $AB$ can slide on two parallel thick metallic rails separated by a distance $l$. A resistance $R$ and an inductance $L$ are connected to the rails as shown in the figure. A long straight wire, carrying a constant current $I_0$ is placed in the plane of the rails as shown. The bar $AB$ is held at rest at a distance $x_0$ from the long wire. At $t = 0$, it made to slide on the rails away from the wire. Answer the following questions.
(a) Find a relation among $i$, $\frac{di}{dt}$ and $\frac{d\phi}{dt}$, where $i$ is the current in the circuit and $\phi$ is the flux of the magnetic field due to the long wire through the circuit.
(b) It is observed that at time $t = T$, the metal bar $AB$ is at a distance of $2x_0$ from the long wire and the resistance $R$ carries a current $i_1$. Obtain an expression for the net charge that has flown through resistance $R$ from $t = 0$ to $t = T$.
(c) The bar is suddenly stopped at time $T$. The current through resistance $R$ is found to be $i_1/4$ at time $2T$. Find the value of $L/R$ in terms of the other given quantities.

Sigma Percentile · Accepted Answer

### The Setup: A Moving Bar in a Magnetic Field

Imagine a long, straight wire carrying a steady current $I_0$ upwards. According to Ampere's Law and the right-hand rule, this wire generates a magnetic field that points directly into the page on its right side. The magnitude of this field at any distance $x$ is given by $B = \frac{\mu_0 I_0}{2\pi x}$.

In this magnetic field, we have a conducting bar $AB$ resting on two parallel rails. The rails are connected to a resistor $R$ and an inductor $L$, forming a complete circuit. When the bar starts sliding away from the wire, the area of the loop increases, and the magnetic field it sweeps through changes. This changing magnetic flux induces an electromotive force (EMF) in the loop, driving a current $i$.

### Part (a): The Master Equation

To understand the dynamics of this circuit, we apply Kirchhoff's Voltage Law (KVL). The induced EMF $e$ acts as our voltage source. As the current flows, there is a voltage drop across the resistor ($iR$) and the inductor ($L\frac{di}{dt}$).

$$e - iR - L\frac{di}{dt} = 0$$

From Faraday's Law of Induction, the magnitude of the induced EMF is exactly the rate of change of magnetic flux, $e = \frac{d\phi}{dt}$. Substituting this into our KVL equation gives us the fundamental relation governing the circuit:

$$\frac{d\phi}{dt} = iR + L\frac{di}{dt}$$

### Part (b): Tracking the Flow of Charge

We are asked to find the net charge $\Delta q$ that flows through the resistor as the bar moves from its initial position $x_0$ to a new position $2x_0$ over a time interval $T$. To do this, we rearrange our master equation by multiplying every term by $dt$:

$$d\phi = iR dt + L di$$

Now, we integrate this equation from $t = 0$ to $t = T$. The integral of current over time ($\int i dt$) is precisely the total charge $\Delta q$. The current starts at $0$ and reaches $i_1$ at time $T$.

$$\Delta \phi = R \Delta q + L i_1$$

To find $\Delta \phi$, we integrate the magnetic field over the area swept by the bar. The area element is $dA = l dx$.

$$\Delta \phi = \int_{x_0}^{2x_0} \frac{\mu_0 I_0}{2\pi x} l dx = \frac{\mu_0 I_0 l}{2\pi} \ln\left(\frac{2x_0}{x_0}ight) = \frac{\mu_0 I_0 l}{2\pi} \ln(2)$$

Substituting this back into our integrated equation, we can easily isolate $\Delta q$:

$$\Delta q = \frac{1}{R} \left[ \frac{\mu_0 I_0 l}{2\pi} \ln(2) - L i_1 ight]$$

### Part (c): The Sudden Halt and Exponential Decay

At exactly time $t = T$, the bar is suddenly stopped. Its velocity drops to zero instantly. Because motional EMF depends on velocity ($e = Bvl$), the induced EMF also vanishes immediately.

Without a voltage source, the circuit transforms into a simple decaying $L-R$ circuit. The inductor, which resists sudden changes in current, will force the current to decay exponentially from its value $i_1$ at time $T$. Let $t'$ be the time elapsed after the bar stops ($t' = t - T$). The current follows the standard decay equation:

$$i(t') = i_1 e^{-t'/	au_L}$$

We are given a crucial piece of information: at time $t = 2T$ (which means $t' = T$), the current has dropped to $i_1/4$. Let's plug this in:

$$\frac{i_1}{4} = i_1 e^{-T/	au_L}$$

Canceling $i_1$ and taking the reciprocal gives $e^{T/	au_L} = 4$. Taking the natural logarithm of both sides yields:

$$\frac{T}{	au_L} = \ln 4$$

Since the time constant $	au_L$ is defined as $L/R$, we arrive at our final elegant result:

$$\frac{L}{R} = \frac{T}{\ln 4}$$

Visualized Solution

$B = \frac{μ _{0} I _{0}}{2 π x}$

$Kirchhoff’s Voltage Law$

$\frac{d ϕ}{d t} = i R + L \frac{d i}{d t}$

$d ϕ = i R d t + L d i$

$Δ ϕ = R Δ q + L i_{1}$

$Motion of the Bar$

$Δ ϕ = \frac{μ _{0} I _{0} l}{2 π} ln (2)$

$Δ q = \frac{1}{R} [\frac{μ _{0} I _{0} l}{2 π} ln (2) - L i_{1}]$

$v = 0 ⟹ e = 0$

$i (t^{'}) = i_{0} e^{- t^{'} / τ_{L}}$

$\frac{i _{1}}{4} = i_{1} e^{- T / τ_{L}}$

$\frac{L}{R} = \frac{T}{l n 4}$

The Sigma Insight: Motional EMF

The Setup

Part (a)

Part (b)

Part (c)

Similar Questions

A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R Ω$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_{1}, I_{2}$ and $I$ are

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B=2πxμ0​I0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kirchhoff’s Voltage Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdϕ​=iR+Ldtdi​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dϕ=iRdt+Ldi

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δϕ=RΔq+Li1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Motion of the Bar

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δϕ=2πμ0​I0​l​ln(2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δq=R1​[2πμ0​I0​l​ln(2)−Li1​]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=0⟹e=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i(t′)=i0​e−t′/τL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4i1​​=i1​e−T/τL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RL​=ln4T​