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Motional emf | Motional Electromotive Force

Motional emf | Motional Electromotive Force :- Consider a conductor of length l moving with velocity v through a uniform magnetic field B, as shown in figure below.

The electrons inside the wire also move with velocity v, and each experiences a magnetic force $\displaystyle \overrightarrow{{{F}_{B}}}=-e\left( \overrightarrow{v}\times \overrightarrow{B} \right)$ of magnitude F = evB. This force pushes the electrons toward one side of the conductor (downward in the figure). As electrons accumulate on that side, the opposite side becomes positively charged, leading to a separation of charges and the formation of an electric field inside the conductor.

The charge carriers keep separating until the electric force F_E = eE exactly balances the magnetic force F_B = evB, reaching equilibrium. This occurs when :

$\displaystyle \left| \overrightarrow{{{F}_{E}}} \right|=\left| \overrightarrow{{{F}_{B}}} \right|$

$\displaystyle \Rightarrow eE=evB$

$\displaystyle \Rightarrow E=vB$ …..(1)

So, we can say that the magnetic force on the charge carriers in a moving conductor generates an electric field inside the conductor.

This electric field, in turn, creates an electric potential difference between the two ends of the moving conductor.

Using the relation between the electric field and the electric potential,

$\displaystyle E=-\frac{dV}{dr}$

$\displaystyle \Rightarrow dV=-\overrightarrow{E}.\overrightarrow{dr}$

Along the length of the moving conductor,

$\displaystyle dV=-\left( \overrightarrow{E} \right).\left( \overrightarrow{dy} \right)=-\left( -vB\widehat{j} \right).(dy\widehat{j})=vBdy$

Integrating,

$\displaystyle \Delta V=\int\limits_{0}^{l}{vBdy}=vB\int\limits_{0}^{l}{dy}=vB\left[ y \right]_{0}^{l}=vB\left[ l-0 \right]$

$\displaystyle \Rightarrow \Delta V=Blv$

Motional emf formula

Thus, moving the wire through a magnetic field induces a potential difference Δ between its ends, which depends on the magnetic field strength and the speed of the conductor.

Vector form of Motional emf | Motional Electromotive Force

At equilibrium, the net force on the electron is zero, so

$\displaystyle \overrightarrow{{{F}_{E}}}+\overrightarrow{{{F}_{B}}}=0$

$\displaystyle \Rightarrow q\overrightarrow{E}+q\left( \overrightarrow{v}\times \overrightarrow{B} \right)=0$

$\displaystyle \Rightarrow \overrightarrow{E}=-\left( \overrightarrow{v}\times \overrightarrow{B} \right)$

Now, because

$\displaystyle dV=-\vec{E}.\overrightarrow{dr}$

Hence

$\displaystyle dV=-\vec{E}.\overrightarrow{dr}=-\left[ -\left( \overrightarrow{v}\times \overrightarrow{B} \right) \right].\overrightarrow{dl}$

$\displaystyle \Rightarrow dV=\left( \overrightarrow{v}\times \overrightarrow{B} \right).\overrightarrow{dl}$

On integrating,

$\displaystyle \Delta V=\int\limits_{\text{0}}^{l}{\left( \overrightarrow{v}\times \overrightarrow{B} \right).\overrightarrow{dl}}=\left( \overrightarrow{v}\times \overrightarrow{B} \right).\int\limits_{\text{0}}^{l}{\overrightarrow{dl}}=\left( \overrightarrow{v}\times \overrightarrow{B} \right).\left[ \overrightarrow{l} \right]_{0}^{l}$

$\displaystyle \therefore \Delta V=\left( \overrightarrow{v}\times \overrightarrow{B} \right).\overrightarrow{l}$ …..(3)

The above equation (3) is the vector representation of Motional emf.

Analogy Between Battery emf and Motional emf

There’s an interesting analogy between the potential difference across a moving conductor and that of a battery. A battery uses a non-electric (chemical) force, often called a charge escalator, to separate positive and negative charges. The emf (ε) of the battery is defined as the work done per unit charge (W/q) to separate these charges. Since this separation is caused by chemical reactions, a battery is said to be a source of chemical emf.

Similarly, a moving conductor develops a potential difference because magnetic forces do the work of separating charges. You can think of it as a kind of “battery” whose emf arises from motion, not chemistry — it produces emf only while it moves and stops doing so when motion ceases.

Motional emf by changing the magnetic flux enclosed by a circuit

Consider a rectangular conducting loop PQRS placed in a uniform magnetic field $\displaystyle \overrightarrow{B}$ , directed perpendicular to the plane of the loop (say, into the page).

MN is a movable conductor of length $l$ , sliding with a constant velocity along conducting rails PQ and RS. Assume that there is no loss of energy due to friction. MQRN forms a closed circuit enclosing an area that changes as MN moves. If at any instant the length RN = x, then the magnetic flux through the loop MQRN is :

$\displaystyle {{\phi }_{B}}=Blx$

Since x is changing with time, the rate of change of flux Φ_B will induce an emf given by :

$\displaystyle \varepsilon =-\frac{d{{\phi }_{B}}}{dt}=-\frac{d\left( Blx \right)}{dt}=-Bl\frac{dx}{dt}$

Here $\displaystyle \frac{dx}{dt}=v$ , velocity of the conductor MN, hence induced emf :

$\displaystyle \varepsilon =-Blv$ …..(4)

The negative sign indicates that the induced current flows in a direction that produces a magnetic field out of the page, opposing the increase of flux into the page (Lenz’s law). The magnitude of the motional emf is Blv.

If the total resistance of the circuit is R, the induced current is given by Ohm’s law as

$\displaystyle I=\frac{\left| \varepsilon \right|}{R}=\frac{Blv}{R}$ …..(5)

Force Required To Move The Conductor

(Motional emf | Motional Electromotive Force)

The moving conductor MN now carries an induced current and lies in a magnetic field. Since a magnetic field exerts a force on a current-carrying wire, the direction of this force can be determined using Fleming’s left-hand rule. In this case, the magnetic force acts towards the left, as shown in the diagram below, and its magnitude is given by F_B = BIl.

This “magnetic drag” will slow and eventually stop the conductor MN unless an equal and opposite external pulling force is applied to keep it moving. Hence using equation (4), force required to pull the wire with a constant speed v is given by :

$\displaystyle \left| {{\overrightarrow{F}}_{ext}} \right|=\left| \overrightarrow{{{F}_{B}}} \right|=BIl=B\left( \frac{Blv}{R} \right)l=\frac{{{B}^{2}}{{l}^{2}}v}{R}$ …..(6)

Energy Considerations

To pull the conductor MN along the track, an external work is required. To understand what happens to the energy transferred to the conductor and whether energy is conserved, it is easier to think in terms of power rather than work (If you try to track total energy directly, it becomes messy because energy is constantly changing from one form to another and spread over time). Power represents the rate at which work is done on the conductor.

The mechanical power delivered to the conductor by the force applied on it :

$\displaystyle {{P}_{input}}={{F}_{ext}}v=\frac{{{B}^{2}}{{l}^{2}}v}{R}\times v=\frac{{{B}^{2}}{{l}^{2}}{{v}^{2}}}{R}$ …..(7)

The electrical power dissipated in the circuit :

$\displaystyle {{P}_{electrical}}={{I}^{2}}R={{\left( \frac{Blv}{R} \right)}^{2}}R=\frac{{{B}^{2}}{{l}^{2}}{{v}^{2}}}{R}$ …..(8)

From equations (7) and (8) it is clear that the mechanical power delivered by the pulling force is fully converted into electrical power in the circuit, showing that both expressions are equal and energy is conserved.

Potential Difference Along A Rotating Rod

Consider a conducting rod of length rotating with a uniform angular velocity about one of its ends (say end ) in a uniform magnetic field directed perpendicular to the plane of rotation (into the plane of the paper).

Using Fleming’s left-hand rule, the electrons in the rod experience a magnetic force directed towards the pivot. As a result, the outer end of the rod becomes positive, while the inner end (the pivot) becomes negative.

Method I

Since the rod is rotating with an angular velocity ω, the linear speed of any element of the rod at a distance r from the pivot is v = ωr. Using equation (1), the electric field intensity at this point is :

The direction of E along the rod is towards the pivot, and the magnitude of E increases outward from the pivot. On integrating E with respect to r, outward from the center, the potential difference between the ends of the rod is :

$\displaystyle \Delta V={{V}_{tip}}-{{V}_{pivot}}=-\int\limits_{0}^{L}{\overrightarrow{E}.\overrightarrow{dr}}=-\int\limits_{0}^{L}{\left( \omega rB \right)}\left( dr \right)\left( cos180{}^\circ \right)$

$\displaystyle \Rightarrow \Delta V=-\int\limits_{0}^{L}{\left( \omega rB \right)}\left( dr \right)\left( -1 \right)=\omega B\int\limits_{0}^{L}{rdr=}\omega B\left[ \frac{{{r}^{2}}}{2} \right]_{0}^{L}$

Hence generated emf ,

$\displaystyle \therefore \varepsilon =\Delta V=\frac{1}{2}B\omega {{L}^{2}}$ …..(10)

Method II

Consider a small element of length dr at a distance r from the pivot. The magnitude of the emf induced across this element, moving at right angles to the magnetic field, is given by

dε = Bvdr = Bdr

On integrating dε along the length of the rod, we obtain the total induced emf as :

$\displaystyle \varepsilon =\int\limits_{0}^{L}{B\omega rdr}=B\omega \int\limits_{0}^{L}{rdr=}\omega B\left[ \frac{{{r}^{2}}}{2} \right]_{0}^{L}=\frac{1}{2}B\omega {{L}^{2}}$

Example 1.

NCERT Example 6.7

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field HE at a place. If H_E = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? Note that 1 G = 10^–4 T.

Solution :

$\displaystyle \varepsilon =\frac{1}{2}B\omega {{L}^{2}}=\frac{1}{2}B\left( 2\pi \nu \right){{L}^{2}}=\frac{1}{2}\left( 0.4\times {{10}^{-4}} \right)\times \left( 2\pi \times \frac{120}{60} \right){{\left( 0.5 \right)}^{2}}$

$\displaystyle \therefore \varepsilon =6.28\times {{10}^{-5}}Volt$

The number of spokes in the rotating wheel does not affect the total induced emf because all the spokes are connected between the same inner rim (hub) and outer rim. This means the emfs generated across different spokes are in parallel, and hence the potential difference between the rim and the axle remains the same regardless of how many spokes are present.