JEE Advanced 1988

LEVELJEE Advanced

Animated Solution for Physics - Electromagnetic Induction: Two long parallel horizontal rails, a distance $d$ apart and each having a resistance $λ$ per unit length, are joined at one end by a resistance $R$ . A perfectly conducting rod $M N$ of mass $m$ is free to slide along the rails without friction (see figure). There is a uniform magnetic field of induction $B$ normal to the plane of the paper and directed into the paper. A variable force $F$ is applied to the rod $M N$ such that, as the rod moves, a constant current $i$ flows through $R$ . (i) Find the velocity of the rod and the applied force $F$ as functions of the distance $x$ of the rod from $R$ . (ii) What fraction of the work done per second is converted into heat.

Q: Why is the resistance $R + 2\lambda x$ and not just $R + \lambda x$?

Because the current travels down one rail of length $x$ and returns through the other rail of length $x$. Both rails contribute to the resistance.

Q: Why do we need to differentiate velocity to find acceleration? Can't we just use $F = ma$?

We do use $F_{\text{net}} = ma$, but the velocity is not constant. Since the current is constant and the resistance increases with $x$, the velocity must also increase with $x$ to generate enough motional EMF. Thus, the rod is accelerating.

Q: Why is the magnetic force $idB$ and not $ixB$?

The magnetic force acts on the current-carrying rod $MN$. The length of the rod perpendicular to the magnetic field is $d$, the distance between the rails.

Q: Where does the rest of the power go if not all of it is converted to heat?

The applied mechanical power $Fv$ is split into two parts: Joule heating ($i^2 R_{\text{net}}$) and the rate of change of the rod's kinetic energy ($\frac{d}{dt}(\frac{1}{2}mv^2) = mva$).

Q: Why is the current constant?

The problem states that a *variable* force $F$ is applied *such that* the current remains constant. The force is deliberately adjusted to maintain this condition.

Visualized Solution

$R_{net} = R + 2 λ x$

Total resistance of the circuit as a function of distance $x$ from resistance $R$ is:
$R_{net} = R + 2 λ x$

$i = \frac{e}{R _{net}}$

Let $v$ be the velocity of rod at this instant, then motional emf induced across the rod is $e = B v d$ .
Current $i = \frac{e}{R _{net}} = \frac{B v d}{R + 2 λ x}$

$v = \frac{( R + 2 λ x ) i}{B d}$

Rearranging for velocity $v$ :
$v = \frac{( R + 2 λ x ) i}{B d}$

$a = \frac{d v}{d t}$

Differentiating velocity with respect to time to find acceleration:
$a = \frac{d v}{d t} = \frac{d}{d t} [\frac{i ( R + 2 λ x )}{B d}]$
$a = \frac{2 λi}{B d} \frac{d x}{d t}$

$F_{net} = ma$

Since $\frac{d x}{d t} = v = \frac{( R + 2 λ x ) i}{B d}$ :
$a = \frac{2 λi}{B d} [\frac{( R + 2 λ x ) i}{B d}] = \frac{2 λ i ^{2} ( R + 2 λ x )}{B ^{2} d ^{2}}$
Net force on the rod:
$F_{net} = ma = \frac{2 λm i ^{2} ( R + 2 λ x )}{B ^{2} d ^{2}}$

$F = F_{net} + F_{m}$

The net force is equal to $F - F_{m}$ , where $F_{m} = i d B$ is the magnetic force.
$F = F_{net} + F_{m}$
$F = \frac{2 λm i ^{2}}{B ^{2} d ^{2}} (R + 2 λ x) + i d B$

$P_{input} = F v$

Work done by $F$ per second (input power):
$P_{input} = F v = [\frac{2 λm i ^{2}}{B ^{2} d ^{2}} (R + 2 λ x) + i d B] [\frac{i ( R + 2 λ x )}{B d}]$

$P_{heat} = i^{2} R_{net}$

Heat produced per second in the circuit:
$P_{heat} = i^{2} R_{net} = i^{2} (R + 2 λ x)$

$f = \frac{P _{heat}}{P _{input}}$

The desired fraction is:
$f = \frac{P _{heat}}{P _{input}} = \frac{i ^{2} ( R + 2 λ x )}{F v}$
$f = [1 + \frac{2 λmi}{B ^{3} d ^{3}} (R + 2 λ x)]^{- 1}$

$What if F = constant?$

Consider: If $F = constant$ , find $v (t)$ .

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The Sigma Insight: Motional EMF

Solution Diagram

This problem is a beautiful interplay of electromagnetism, kinematics, and energy conservation. It challenges us to think dynamically because the very circuit we are analyzing is changing its physical dimensions as time progresses.

Analyzing the Setup

Imagine the rod sliding along the rails. As it moves further away from the resistor

R

, the length of the rails included in the active circuit increases. Since the rails have a resistance of

λ

per unit length, the total resistance of the circuit is not constant. It is a function of the rod's position

x

R_{net} = R + 2 λ x

We multiply by

2

because the current must travel down one rail and return through the other.

The Master Equation for Velocity

As the rod moves with velocity

v

through the uniform magnetic field

B

, it generates a motional EMF,

e = B v d

. According to Ohm's law, the current

i

in the circuit is this EMF divided by the total resistance:

i = \frac{B v d}{R + 2 λ x}

The problem states a crucial constraint: the current

i

is constant. This means that as the resistance

R + 2 λ x

increases, the velocity

v

must also increase proportionally to maintain the constant current. Rearranging for

v

, we get:

v = \frac{i ( R + 2 λ x )}{B d}

Calculating the Applied Force

Because the velocity

v

is increasing with

x

, the rod is accelerating. We can find this acceleration by differentiating

v

with respect to time using the chain rule:

a = \frac{d v}{d t} = \frac{d}{d t} [\frac{i ( R + 2 λ x )}{B d}] = \frac{2 λi}{B d} \frac{d x}{d t}

Since

\frac{d x}{d t}

is simply the velocity

v

, we substitute our expression for

v

back into the equation:

a = \frac{2 λi}{B d} [\frac{i ( R + 2 λ x )}{B d}] = \frac{2 λ i ^{2} ( R + 2 λ x )}{B ^{2} d ^{2}}

According to Newton's second law, the net force required to produce this acceleration is

F_{net} = ma

. However, the applied force

F

must not only provide this net force but also overcome the opposing magnetic force

F_{m} = i d B

acting on the current-carrying rod. Therefore:

F = F_{net} + F_{m} = \frac{2 λm i ^{2}}{B ^{2} d ^{2}} (R + 2 λ x) + i d B

Energy Conversion

Work to Heat

For the second part, we need to find the fraction of the mechanical work done per second that is converted into Joule heating. The work done per second is the power input from the applied force: