A conducting wire XY of mass m and negligible resistance can slide smoothly, staying perpendicular to two long

NCERT Exemplar · Q19

Q. A conducting wire XY of mass m and negligible resistance can slide smoothly, staying perpendicular to two long parallel horizontal conducting rails — one running along AB (the upper rail) and one along CD (the lower rail). The two rails are joined at their left end A-C through a resistance R, while the rail segments AB and CD are themselves perfect conductors. The perpendicular separation between the two rails is l, and the instantaneous distance of the wire XY from the closed (resistor) end is x(t). A uniform magnetic field B = B(t) points out of the plane of the rails.

(i) Write the equation for the acceleration of the wire XY.

(ii) If B is independent of time, obtain v(t), assuming v(0) = u₀.

(iii) For part (ii), show that the decrease in the kinetic energy of XY equals the heat dissipated in R.

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Concept understanding — Motional Emf

Motional Emf

When a conductor moves through a magnetic field, the free charges inside it experience a magnetic force. This force pushes the charges along the conductor, setting up a potential difference across its ends — a motional emf. It is electromagnetic induction viewed from a moving conductor rather than from a changing field.

Origin: the Lorentz force

Consider a straight rod of length $l$ moving with velocity $\vec{v}$ perpendicular to a uniform field $\vec{B}$ . Each free charge $q$ feels a force

\vec{F} = q\,(\vec{v}\times\vec{B})

of magnitude $qvB$ directed along the rod. Charges pile up at the ends until the electric field they create balances the magnetic force. The resulting potential difference — the motional emf — is

\varepsilon = Bvl

Consistency with Faraday's law

Let the rod of length $l$ slide along parallel conducting rails, sweeping out a distance $x$ . The enclosed area is $A = lx$ , so the flux is $\Phi = Blx$ . Then

\varepsilon = -\frac{d\Phi}{dt} = -Bl\frac{dx}{dt} = -Blv

The magnitude $Bvl$ matches the Lorentz-force result, showing that the flux rule and the force picture agree.

Worked example

A rod of length $0.4\,\text{m}$ moves at $5\,\text{m/s}$ perpendicular to a field of $0.5\,\text{T}$ :

\varepsilon = Bvl = 0.5\times5\times0.4 = 1\,\text{V}

If the circuit resistance is $2\,\Omega$ , the induced current is $I = \varepsilon/R = 0.5\,\text{A}$ .

Force and energy …

Why this formula?

Motional EMF: Why the Formula Holds

Let's build this from first principles — understanding the why before the formula.

The Core Idea

Motional EMF arises when a conductor moves through a magnetic field. The key insight: moving charges in a magnetic field experience a magnetic force, which acts like a battery pushing charges around the conductor.

Step 1: The Force on a Moving Charge

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ feels the Lorentz magnetic force:

\vec{F}_m = q (\vec{v} \times \vec{B})

This force is perpendicular to both velocity and magnetic field.

Step 2: What Happens Inside a Moving Conductor

Consider a straight metal rod of length $L$ moving with constant velocity $\vec{v}$ perpendicular to a uniform magnetic field $\vec{B}$ (pointing into the page).

Free electrons in the rod are moving with the rod at velocity $\vec{v}$ .
Each electron experiences a magnetic force:

\vec{F}_m = -e (\vec{v} \times \vec{B})

(negative sign because electron charge is $-e$ )

This force pushes electrons along the rod — say, toward one end.

Step 3: Charge Separation Creates an Electric Field

As electrons accumulate at one end, that end becomes negatively charged, leaving the other end positively charged.

This charge separation creates an internal electric field $\vec{E}$ inside the rod, pointing from positive to negative end.
The electric field exerts an opposing force on the electrons:

\vec{F}_e = -e \vec{E}

Step 4: Equilibrium — The "Battery" is Formed

Charge keeps moving until the electric force balances the magnetic force:

\vec{F}_e + \vec{F}_m = 0

-e \vec{E} - e (\vec{v} \times \vec{B}) = 0

\vec{E} = - (\vec{v} \times \vec{B})

Magnitude-wise (for perpendicular $\vec{v}$ and $\vec{B}$ ):

E = vB

Step 5: From Electric Field to EMF

The motional EMF $\mathcal{E}$ is the work done per unit charge to move a test charge from one end to the other:

\mathcal{E} = \int_{\text{negative}}^{\text{positive}} \vec{E} \cdot d\vec{l}

For a uniform field along the rod of length $L$ :

\mathcal{E} = E \cdot L = vBL

The Key Formula

Motional EMF for a straight conductor moving perpendicular to $\vec{B}$ :

$\boxed{\mathcal{E} = B L v}$

Why This Makes Physical Sense

Quantity	Role
$B$	Stronger magnetic field → larger force on charges
$L$	Longer conductor → more charge separation possible

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The moving rod XY generates an emf; the induced current in field B produces a retarding force $F=-BIl$ on the rod, giving the acceleration equation. With B constant the rod's speed decays exponentially, $v(t)=u_0e^{-B^2l^2t/mR}$ , and the kinetic energy it loses is dissipated exactly as $I^2R$ heat.

(i) Equation of motion

The flux through the circuit is $\phi=B(t)\,l\,x$ . The emf is

\varepsilon=-\frac{d\phi}{dt}=-l\left(B\frac{dx}{dt}+x\frac{dB}{dt}\right)=-l\left(Bv+x\frac{dB}{dt}\right).

The current is $I=\dfrac{|\varepsilon|}{R}=\dfrac{l}{R}\left(Bv+x\dfrac{dB}{dt}\right)$ , and the force on the current-carrying rod ( $F=BIl$ ) opposes the motion. Newton's second law:

m\frac{dv}{dt}=-BIl=-\frac{Bl^2}{R}\left(Bv+x\frac{dB}{dt}\right).

(ii) B independent of time

Put $\dfrac{dB}{dt}=0$ :

m\frac{dv}{dt}=-\frac{B^2l^2}{R}\,v\;\Rightarrow\;\frac{dv}{v}=-\frac{B^2l^2}{mR}\,dt.

Integrating with $v(0)=u_0$ :

v(t)=u_0\,e^{-B^2l^2t/(mR)}.

(iii) Energy balance

Kinetic energy lost as $t:0\to\infty$ (the rod eventually stops): …

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Concept understanding — Motional Emf

Motional Emf

Origin: the Lorentz force

Consistency with Faraday's law

Worked example

Force and energy …

Why this formula?

Motional EMF: Why the Formula Holds

The Core Idea

Step 1: The Force on a Moving Charge

Step 2: What Happens Inside a Moving Conductor

Step 3: Charge Separation Creates an Electric Field

Step 4: Equilibrium — The "Battery" is Formed

Step 5: From Electric Field to EMF

The Key Formula

Why This Makes Physical Sense

(i) Equation of motion

(ii) B independent of time

(iii) Energy balance

Method: Setting Up and Solving the Equation of Motion for an Electromagnetically Damped Conductor

Steps

(i) Equation of motion

(ii) B independent of time

(iii) Energy balance

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