IIT Cut Off for BTech Industrial Chemistry: Check Previous Year Cutoff

# Motional Electromotive Force - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Motional Electromotive force(IV) is considered one the most difficult concept.

• Motional Electromotive force(I), Motional Electromotive force(II), Energy consideration in Motional Emf, Motional Electromotive force(III) is considered one of the most asked concept.

• 68 Questions around this concept.

## Solve by difficulty

A rectangular loop has a sliding connector PQ of length $\dpi{100} l$ and resistance $\dpi{100} R\: \Omega$ and it is moving with a speed $\dpi{100} \upsilon$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $\dpi{100} I_{1},I_{2}\: and \: I_{3}$ are

A conducting square loop of side L and resistance R moves in its plane with a uniform velocity perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure.

The induced emf is

A wire of length $1 m$ moving with velocity $8 m/s$ at right angles to a magnetic field of
$2 T$. The magnitude of induced emf, between the ends of the wire will be :

An emf of $0.08 \mathrm{~V}$is induced in a metal rod of length $10 \mathrm{~cm}$ held normal to a uniform magnetic field of $0.4 \mathrm{~T}$,  when moves with a velocity of:

A boat is moving due east in a region where the earth’s magnetic field is 5.0 × 10-5 N A-1 m-1 due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 m s-1, the magnitude of the induced emf in the wire of the aerial is

A metallic rod of length 'I' is tied to a string of length 2I and made to rotate with angular speed on a horizontal table with one end of the string fixed.If there is a vertical magnetic field 'B' in the region,the e.m.f. induced across the ends of the rod is :

In an AC generator, a coil with N turns, all of the same area A and total resistance R , rotates with frequency in a magnetic field B . The maximum value of emf generated in the coil is

##### Amity University, Noida B.Tech Admissions 2024

Asia's Only University with the Highest US & UK Accreditation

Ranked #52 among universities in India by NIRF | Highest CTC 50 LPA | 100% Placements | Last Date to Apply: 27th June

## Concepts Covered - 5

Motional Electromotive force(I)

If a conducting rod of length $l$ is moving with a uniform velocity $\vec{V}$ perpendicular to the region of the uniform magnetic field ($\vec{B}$) which directed into the plane of the paper as shown in the below figure.

Then the magnetic force on +ve charges is given by $\vec{F}_{B}=q (\vec{v} \times \vec{B})=e (\vec{v} \times \vec{B})$ toward side b.

And similarly the magnetic force on -ve charges is given by $\vec{F}_{B}=q (\vec{v} \times \vec{B})=e (\vec{v} \times \vec{B})$ toward side a.

So positive and negative charges will accommodate at side b and side a respectively. This will create an electric field having direction from b to a. And electric force due to this field on charges will be given as $\vec{F}_{E}=q \vec{E}$

Applying Equilibrium condition between electric and magnetic force

$F_{E}=F_{B} \quad \Rightarrow q E=q v B \quad \Rightarrow \quad E=v B$

So Potential difference induced between endpoints of the rod is given by

$V_{a b} \equiv V_{b}-V_{a}=E L \quad \Rightarrow V_{a b}=v B L$

this Potential difference ($V_{a b}$) is known as motional emf.

So Motional EMF is given by

$\varepsilon = Blv$
where

$B\rightarrow$ magnetic field

$l\rightarrow$ length of conducting

$v\rightarrow$ the velocity of rod perpendicular to a uniform magnetic field.

Motional Electromotive force(II)

If conducting PQ rod moves on two parallel conducting rails as shown in below  figure

and we wanted to find motional emf of the moving rod

Method I-

As magnetic flux is given by $\phi =B.A$

So initial flux passing through PQRS is given by $\phi =B.A=B(l.x)$

And when rod starts moving this flux will change then the change in flux ix given as $\varepsilon=-\frac{d \phi}{d t}=-\frac{d}{d t}(Blx)=-Bl\frac{dx}{dt}=-Bl(-v)=Blv$

So the motional emf is given as    $\varepsilon= Blv$

Method II-

Due to the motion of the rod +ve and -ve charges of the rod will start to move towards point Q and P respectively.

Then the magnetic force on +ve charges is given by $\vec{F}_{B}=q (\vec{v} \times \vec{B})=e (\vec{v} \times \vec{B})$ toward   Q.

And similarly, the magnetic force on -ve charges is given by $\vec{F}_{B}=q (\vec{v} \times \vec{B})=e (\vec{v} \times \vec{B})$ toward  P.

So the work  done by the magnetic force to move the +ve charge from P to Q is given by $W=\vec{F}_{B}.l=q (\vec{v} \times \vec{B}).l= qvBl$

So potential difference across PQ is given as $\Delta V=V_{PQ}=\frac{W}{q}=Blv$

So the motional emf is given as    $\varepsilon= Blv$

Energy consideration in Motional Emf

As we learn for the above figure Motional EMF is given by

$\varepsilon = Blv$
where

$B\rightarrow$ magnetic field

$l\rightarrow$ length of conducting

$v\rightarrow$ the velocity of rod perpendicular to a uniform magnetic field.

So now we want to find whether the law of conservation is applicable for the motional emf or not?

So Induced Current in the conducting rod is given as   $I= \frac{\varepsilon }{r}= \frac{Blv}{r}$

where r is the resistance of the rod

And assuming resistance of other arms (i.e PS,SR,RQ) is negligible.

Magnetic force on conducting rod is given as

$F=IlB=B\left ( \frac{Blv}{r} \right )l$

$F=\frac{B^{2}vl^{2}}{r}$

The power dissipated in moving the conducting rod -

$P_{mech}=P_{ext}=F\cdot v=\left ( \frac{B^{2}vl^{2}}{r} \right )\cdot v$

$P_{mech}=P_{ext}= \frac{B^{2}l^{2}v^{2}}{r}$

Electric Power or the  rate of heat dissipation across the resistance is given as

$P_{E}=I^{2}r=\left ( \frac{Blv}{r} \right )^{2}\cdot r= \frac{B^{2}l^{2}v^{2}}{r}$

Since $P_{mech}=P_{E}$  So we can say that the principle of conservation of energy is applicable for the motional emf.

Motional Electromotive force(III)

General Case-

Motional emf when $\vec{B} \ and \ \vec{V} \ and \ \vec{l}$ are at some angle with each other as shown in the below figure.

\begin{aligned} \text { Then At steady state, } &\left|F_{e}\right|=\left|F_{m}\right| \\ & \Rightarrow F_{e}=-F_{m} \\ \Rightarrow & e \vec{E}=-\ell(\vec{V} \times \vec{B}) \\ \Rightarrow \vec{E} &=-(\vec{V} \times \vec{B}) \end{aligned}

\begin{aligned} And \ Poential \ difference &=d v=-\vec{E} \cdot \overrightarrow{d l} \\ & \Rightarrow d v=\int(\vec{V} \times \vec{B}) \cdot \vec{dl} \\ \Rightarrow \Delta v=(\vec{V} \times \vec{B}) \cdot \vec{l} \\ \\ \Rightarrow \varepsilon =(\vec{V} \times \vec{B}) \cdot \vec{l} & \end{aligned}

For example-

• If the rod is moving by making an angle $\theta$ with the direction of the magnetic field or length as shown in the below figure.

then Induced emf  $\Rightarrow \varepsilon =BlVsin\theta$

$\\ \Delta v = potential \ \ difference\\ B = Magnetic \ field \\ V = velocity \ of\ the \ rod$

Motional Electromotive force(IV)

Motional E.m.f due to rotational motion-

If a conducting rod  PQ  is rotating with angular velocity  $\omega$ about its one  end (Q) in a uniform magnetic field as shown in the below figure.

then   $\varepsilon =\frac{1}{2}Bl^{2}\omega =Bl^{2}\pi\nu$

where

$\nu=\frac{\omega }{2\pi }=\frac{1}{T}\rightarrow f\! r\! equency$

$T \rightarrow Time\; period$

Similarly

• For  Cycle wheel rotating with angular velocity $\omega$ about O.

$\varepsilon = \frac{1}{2}Bwl^{2}$

• For Metal Disc

$\varepsilon = \frac{1}{2}Bwr^{2}$

## Study it with Videos

Motional Electromotive force(I)
Motional Electromotive force(II)
Energy consideration in Motional Emf

"Stay in the loop. Receive exam news, study resources, and expert advice!"