Motional Electromotive Force - Practice Questions & MCQ

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})_{\text {toward side } \mathrm{b} .}
$

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

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 \Rightarrow q E=q v B \Rightarrow 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=B l v
$

 where 

$B \rightarrow$ magnetic field
$l \rightarrow$ length of conducting
$v \rightarrow$ the velocity of rod perpendicular to a uniform magnetic field.

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 $\phi=B \cdot A=B(l . x)$
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}(B l x)=-B l \frac{d x}{d t}=-B l(-v)=B l v
$

So the motional emf is given as $\varepsilon=B l v$
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 $\mathbf{Q}$.

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 $W=\vec{F}_B \cdot l=q(\vec{v} \times \vec{B}) \cdot l=q v B l$

So potential difference across PQ is given as $\Delta V=V_{P Q}=\frac{W}{q}=B l v$
So the motional emf is given as $\varepsilon=B l v$

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

$
\varepsilon=B l v
$

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 $I=\frac{\varepsilon}{r}=\frac{B l v}{r}$
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

$
\begin{aligned}
& F=I l B=B\left(\frac{B l v}{r}\right) l \\
& F=\frac{B^2 v l^2}{r}
\end{aligned}
$

The power dissipated in moving the conducting rod -

$
\begin{aligned}
& P_{\text {mech }}=P_{e x t}=F \cdot v=\left(\frac{B^2 v l^2}{r}\right) \cdot v \\
& P_{\text {mech }}=P_{e x t}=\frac{B^2 l^2 v^2}{r}
\end{aligned}
$

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

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

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

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.

Then At steady state,

$
\begin{aligned}
& \text { tate, }\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}
& \text { And Poential difference }=d v=-\vec{E} \cdot \overrightarrow{d l} \\
& \qquad \begin{aligned}
\Rightarrow d v=\int(\vec{V} \times \vec{B}) \cdot \overrightarrow{d l} \\
\Rightarrow \Delta v=(\vec{V} \times \vec{B}) \cdot \vec{l}
\end{aligned} \\
& \qquad \begin{array}{l}
\Rightarrow \quad(\vec{V} \times \vec{B}) \cdot \vec{l}
\end{array}
\end{aligned}
$
 

For example-

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

$\begin{aligned} & \text { then Induced emf } \Rightarrow \varepsilon=B l V \sin \theta \\ & \Delta v=\text { potential difference } \\ & B=\text { Magnetic field } \\ & V=\text { velocity of the rod }\end{aligned}$

Motional E.m.f due to rotational motion-

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

then $\varepsilon=\frac{1}{2} B l^2 \omega=B l^2 \pi \nu$
where
$\nu=\frac{\omega}{2 \pi}=\frac{1}{T} \rightarrow$ frequency
$T \rightarrow$ Time period
Similarly

• For Cycle wheel rotating with angular velocity $\omega$ about 0.

$\varepsilon=\frac{1}{2} B w l^2$

• For Metal Disc

$\varepsilon=\frac{1}{2} B w r^2$