Answer
64.8k+ views
Hint: Note the direction of current for each of the loops and then find the highest and lowest potential points for each of the rings. Find the potential difference between the highest and lowest points for each of the rings from the given magnetic field and radius. The potential of the point on the surface is equal for both of the rings since they are in the same plane.
Find the potential difference between the highest points of the rings from each potential difference of these two rings.
Formula used:
If $A$ is the highest potential point and \[E\] is the lowest potential point of the ring of radius \[r\] moving with a speed $2v$,
The potential difference between two points, ${V_{AE}} = B(2r)(2v)$
and, if $C$ is the highest potential point and $D$ is the lowest potential point, of the ring of radius $3r$ moving with a speed $v$,
The potential difference between two points, ${V_{CD}} = B(2 \times 3r)(v)$
Where, $B$ is the magnetic field acting perpendicular to the plane of the rings.
The potential difference, $V = {V_{CD}} - ( - {V_{AE}}) = {V_{CD}} + {V_{AE}}$
Complete step by step solution:
For the first ring, the direction of velocity is towards the left side. So if we apply Fleming’s right hand rule, the current is flowing from point $A$ to \[E\]. Hence, $A$ is the highest potential point and \[E\] is the lowest potential point of the ring.
Given, \[r\]is the radius of the ring and it is moving with a speed $2v$
$B$ is the magnetic field acting perpendicular to the plane of the rings.
$\therefore $ The potential difference between two points, ${V_{AE}} = B(2r)(2v)$
$ \Rightarrow {V_{AE}} = 4Brv$
For the second ring, the direction of velocity is towards the right side. So if we apply Fleming’s right-hand rule, the current is flowing from point $C$ to \[D\]. Hence, $C$ is the highest potential point and $D$ is the lowest potential point Given, \[3r\] is the radius of the ring and it is moving with a speed $v$
$\therefore $ The potential difference between two points, ${V_{CD}} = B(2 \times 3r)(v)$
$ \Rightarrow {V_{CD}} = 6Brv$
The potential difference ,$V = {V_{CD}} - ( - {V_{AE}}) = {V_{CD}} + {V_{AE}}$
$\therefore V = 6Brv + 4Brv = 10Brv$ .
Hence, the right answer is in option \[{\text{D}}{\text{. }}10{\text{ Brv}}\].
Note: The formula used for calculating the potential difference is coming from the equation of induced current in a straight conductor in motion, where the induced current $I = \dfrac{{Blv}}{R}$, $B$ is the magnetic field, $v$ is the motion, $l$ is the length of the straight conductor and $R$ is the resistance.
$ \Rightarrow IR = Blv$
$ \Rightarrow e = Blv$
$e$ is the electromotive force of the circuit, $e = IR$ [Ohm’s law]
In the above problem, the length of the straight conductor $l = 2r$ i.e the length is equal to the diameter of the ring.
Find the potential difference between the highest points of the rings from each potential difference of these two rings.
Formula used:
If $A$ is the highest potential point and \[E\] is the lowest potential point of the ring of radius \[r\] moving with a speed $2v$,
The potential difference between two points, ${V_{AE}} = B(2r)(2v)$
and, if $C$ is the highest potential point and $D$ is the lowest potential point, of the ring of radius $3r$ moving with a speed $v$,
The potential difference between two points, ${V_{CD}} = B(2 \times 3r)(v)$
Where, $B$ is the magnetic field acting perpendicular to the plane of the rings.
The potential difference, $V = {V_{CD}} - ( - {V_{AE}}) = {V_{CD}} + {V_{AE}}$
Complete step by step solution:
For the first ring, the direction of velocity is towards the left side. So if we apply Fleming’s right hand rule, the current is flowing from point $A$ to \[E\]. Hence, $A$ is the highest potential point and \[E\] is the lowest potential point of the ring.
Given, \[r\]is the radius of the ring and it is moving with a speed $2v$
$B$ is the magnetic field acting perpendicular to the plane of the rings.
$\therefore $ The potential difference between two points, ${V_{AE}} = B(2r)(2v)$
$ \Rightarrow {V_{AE}} = 4Brv$
For the second ring, the direction of velocity is towards the right side. So if we apply Fleming’s right-hand rule, the current is flowing from point $C$ to \[D\]. Hence, $C$ is the highest potential point and $D$ is the lowest potential point Given, \[3r\] is the radius of the ring and it is moving with a speed $v$
$\therefore $ The potential difference between two points, ${V_{CD}} = B(2 \times 3r)(v)$
$ \Rightarrow {V_{CD}} = 6Brv$
The potential difference ,$V = {V_{CD}} - ( - {V_{AE}}) = {V_{CD}} + {V_{AE}}$
$\therefore V = 6Brv + 4Brv = 10Brv$ .
Hence, the right answer is in option \[{\text{D}}{\text{. }}10{\text{ Brv}}\].
Note: The formula used for calculating the potential difference is coming from the equation of induced current in a straight conductor in motion, where the induced current $I = \dfrac{{Blv}}{R}$, $B$ is the magnetic field, $v$ is the motion, $l$ is the length of the straight conductor and $R$ is the resistance.
$ \Rightarrow IR = Blv$
$ \Rightarrow e = Blv$
$e$ is the electromotive force of the circuit, $e = IR$ [Ohm’s law]
In the above problem, the length of the straight conductor $l = 2r$ i.e the length is equal to the diameter of the ring.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)