A 10 cm long perfectly conducting wire PQ is moving, with a velocity 1cm/s on a pair of horizontal r

Q: Why is the velocity of the wire constant? Doesn't the magnetic force slow it down?

Excellent observation! Yes, an induced current creates a magnetic force ($F = ilB$) opposing the motion. However, the problem explicitly states to *assume* the velocity remains constant, implying an external mechanical force is perfectly balancing this magnetic drag.

Q: Why do we use the formula for current growth instead of just $I = V/R$?

Because the circuit contains an inductor! An inductor opposes any change in current. When the switch is closed, the current cannot jump to $V/R$ instantly; it must grow exponentially over time.

Q: What if the rails had resistance?

If the rails had resistance, the total resistance of the circuit would increase as the wire moves further away, making $R$ a function of time ($R(t)$). This would require solving a more complex differential equation!

Q: Why is the motional EMF constant?

Motional EMF is given by $\varepsilon = Blv$. Since the magnetic field ($B$), the length of the wire ($l$), and the velocity ($v$) are all constant, the induced EMF remains perfectly constant, acting like a steady DC battery.

Q: I messed up the calculation. How do I handle the units?

Always convert everything to standard SI units before calculating! Length to meters ($10 \text{ cm} = 0.1 \text{ m}$), velocity to m/s ($1 \text{ cm/s} = 0.01 \text{ m/s}$), time to seconds ($1 \text{ ms} = 10^{-3} \text{ s}$), and inductance to Henries ($1 \text{ mH} = 10^{-3} \text{ H}$).

Question

A 10 cm long perfectly conducting wire PQ is moving, with a velocity 1cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R = 1\Omega as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1 T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x 	imes 10^{-3}A, where the value of x is_______. [Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given : e^{-1} = 0.37 , where e is base of the natural logarithm]

Sigma Percentile · Accepted Answer

### The Moving Wire: A Tale of Motional EMF

Imagine a perfectly conducting wire, PQ, sliding smoothly across a pair of frictionless, zero-resistance rails. It's not just moving through empty space; it's slicing through a uniform magnetic field $\mathbf{B}$ that points directly into the plane of your screen.

Whenever a conductor cuts through magnetic field lines, the electrons inside it experience a magnetic Lorentz force. This force pushes the charges to the ends of the wire, creating a potential difference. This phenomenon is known as **Motional EMF**.

According to Faraday's Law of Induction, the magnitude of this induced EMF ($\varepsilon$) for a wire of length $l$ moving with a constant velocity $v$ perpendicular to a magnetic field $B$ is given by the elegant equation:

$$ \varepsilon = Blv $$

Let's plug in the numbers provided in our problem. The magnetic field $B$ is $1 	ext{ T}$. The length of the wire $l$ is $10 	ext{ cm}$, which we must strictly convert to standard SI units as $0.1 	ext{ m}$. The velocity $v$ is $1 	ext{ cm/s}$, or $0.01 	ext{ m/s}$.

Substituting these values, we get:

$$ \varepsilon = (1)(0.1)(0.01) = 10^{-3} 	ext{ V} $$

This is a profound realization. Our simple moving wire has effectively transformed into a tiny, moving DC battery supplying a constant voltage of $10^{-3} 	ext{ V}$!

### The LR Circuit Awakens

Now, let's look at the rest of the setup. The rails are connected to an inductor $L$ and a resistor $R$. The moment we close the switch $S$, we complete the circuit.

If this were a simple resistive circuit, the current would instantly jump to $V/R$. However, we have an inductor in the mix. An inductor is like the "inertia" of an electrical circuit; it strongly opposes any sudden change in current by inducing a back-EMF.

Because of this opposition, the current doesn't reach its maximum value instantly. Instead, it climbs exponentially over time. The mathematical model governing this growth in an LR circuit is:

$$ i(t) = \frac{\varepsilon}{R} \left( 1 - e^{-\frac{Rt}{L}} ight) $$

### The Exponential Climb

To find the current at the specific time $t = 1 	ext{ ms}$, we first need to evaluate the exponent term, $\frac{Rt}{L}$.

Let's gather our parameters: the resistance $R = 1 \, \Omega$, the time $t = 1 	ext{ ms} = 10^{-3} 	ext{ s}$, and the inductance $L = 1 	ext{ mH} = 10^{-3} 	ext{ H}$.

Substituting these into our exponent fraction:

$$ \frac{Rt}{L} = \frac{1 	imes 10^{-3}}{10^{-3}} = 1 $$

The math simplifies beautifully! The $10^{-3}$ terms in the numerator and denominator cancel out perfectly, leaving us with a clean exponent of $1$.

Now, we substitute this back into our main current equation, along with our previously calculated EMF $\varepsilon = 10^{-3} 	ext{ V}$ and resistance $R = 1 \, \Omega$:

$$ i = \frac{10^{-3}}{1} \left( 1 - e^{-1} ight) $$

The problem kindly provides the value of $e^{-1}$ as $0.37$.

$$ i = 10^{-3} (1 - 0.37) $$

$$ i = 0.63 	imes 10^{-3} 	ext{ A} $$

### The Final Verdict

The question asks us to express the current in the form $x 	imes 10^{-3} 	ext{ A}$. By directly comparing our result with this format, we can confidently state that:

$$ x = 0.63 $$

This problem is a beautiful synthesis of two core electromagnetism concepts: the generation of motional EMF and the transient response of an LR circuit. Always remember to watch your units, and never let an exponential equation intimidate you!

Enter Numerical Value:

Visualized Solution

\varepsilon = Blv

\text{Values}

\varepsilon = 10^{-3} \text{ V}

i(t) = \frac{\varepsilon}{R} (1 - e^{-\frac{Rt}{L}})

\frac{Rt}{L}

\frac{Rt}{L} = 1

i = \frac{10^{-3}}{1} (1 - e^{-1})

i = 0.63 \times 10^{-3} \text{ A}

x = 0.63

The Sigma Insight: Motional EMF

The Moving Wire

The LR Circuit Awakens

The Exponential Climb

The Final Verdict

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