Motional Electromotive Force MCQ - Practice Questions with Answers

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

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

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

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 ${\overrightarrow{F}}_E=q\overrightarrow{E}$

Applying Equilibrium condition between electric and magnetic force

$$F_E=F_B\Rightarrow qE=qvB\Rightarrow E=vB$$

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

$$V_{ab}\equiv V_b-V_a=EL\Rightarrow V_{ab}=vBL$$

This potential difference ( $V_{ab}$ ) is known as motional emf.
So Motional EMF is given by

$$\epsilon =Blv$$

where
$B\to$ magnetic field

$l\to$ length of conducting
$v\to$ 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 

So initial flux passing through PQRS is given by 

And when rod starts moving this flux will change then the change in flux ix given as 

So the motional emf is given as    

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  toward   Q. 

And similarly, the magnetic force on -ve charges is given by  toward  P.

So the work  done by the magnetic force to move the +ve charge from P to Q is given by 

So potential difference across PQ is given as  

So the motional emf is given as     

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

 where 

 magnetic field

 length of conducting

 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   

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 

The power dissipated in moving the conducting rod -

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

Since   So we can say that the principle of conservation of energy is applicable for the motional emf.

General Case-

Motional emf when  are at some angle with each other as shown in the below figure.

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.

      then Induced emf  

Motional E.m.f due to rotational motion-

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

then $\epsilon =\frac{1}{2}B{l}^2\omega =B{l}^2\pi \nu$
where

$$\nu =\frac{\omega }{2\pi }=\frac{1}{T}\to \text{ frequency }$$

$T\to$ Time period
Similarly
- For the Cycle wheel rotating with angular velocity $\omega$ about O .

$\epsilon =\frac{1}{2}Bw{l}^2$

• For Metal Disc

$\epsilon =\frac{1}{2}Bw{r}^2$