# 7. Class 12th Physics | EMF Induced in a Rotating Conductor in Uniform Magnetic Field | Ashish Arora

let us study about e m f induced in a rotating

conductor in uniform magnetic field. here we can see, a uniform magnetic field exist

in inward direction, and a straight rod p q is rotating, about its point p as axis of

rotation, with an angular speed omega. here also we can see that this conductor is cutting

magnetic flux in magnetic field so there must be an e m f induced in it. let’s calculate

the e m f by the 2 methods, 1 by motional e m f and other by faraday’s law. in this

situation as, at every point of the rod, the velocity is different, we consider at a distance

x from the center an element of width d x. the velocity of this element is v which we

can write as x omega. this element d x is revolving in a circle of radius x with the

same angular speed omega. now for this element which is of width d x , and velocity v it

is moving, we can write, motional e m f, induced, in element d x is, this we write as d e and

the value will be b v d x, as we know motional e m f is b v l. so here in the element of

width d x motional e m f is b v d x. and by using right hand palm rule we can say that

for this element d x, here its left hand will be high potential, and right end will be low

potential. or we can say this element d x will behave like, a battery of e m f d e and,

several such batteries are connected in series between points p and q. so here in this situation

we can write, total e m f induced, across p and q is, this can be written as e p q which

is integration of d e and this can be given as integration of b, v we can write as x omega

multiplied by d x with limits from zero to l, as l is the length of this rod. so in this

situation this e m f induced across ends p q can be written as b omega as a constant.

and x d x will be integrated as x square by 2 with limits from zero to l. so here induced

e m f across p q is given as, half b omega l square, which itself is a very important

result, and in many cases this can also be used directly. and here by right hand palm

rule we can directly say that point p will be, high potential end, and point q in this

situation will be the low potential end. on the next sheet we’ll continue, and the same

expression, we’ll calculate by using faraday’s law without using the direct result of motional

e m f. let’s continue to understand the same analysis

by using faraday’s law. or this is an alternative analysis to get the same result. here we can

see as the conductor is rotating at an angular speed omega, in time d t we can say the conductor

will reach a position where it is rotated by an angle d-theta in time d t. so we can

write, the angle of rotation, by conductor, in time d t is, this can be written as d-theta

of which the value is omega d t as the angular speed is omega. now in this situation at this

time the conductor swept out this sector area, and we can directly write area of, sector

swept, by conductor, in time d t is, d s we can write as, half of r square d-theta. and

here we can substitute the value of d-theta it is, half r square omega d t. and if we

calculate, magnetic flux cut, in time d t by conductor is, this flux d phi we can write

as b d s. and the area we already calculate so this’ll be half b omega, r square d t.

and by faraday’s law we can directly write, e m f induced, in conductor is e p q which

can be written as mod of d phi by d t, and here on substituting this we’re getting

it as half b omega r square. here we can see the result is same which we’ve obtained,

by using faraday’s law and on the previous sheet, i’ve calculated the same expression,

the same result by using motional e m f result. so we must be careful while analyzing, as

the most important thing here is the e m f induced in the rotating conductor given by

half b omega r square.

## 31 Replies to “7. Class 12th Physics | EMF Induced in a Rotating Conductor in Uniform Magnetic Field | Ashish Arora”

Sir,the second method is wrong,i guess bcoz the rod is not cutting any flux as the magnetic field lines passing through it are consatnt.you are taking the area swept by it rather to take the area enclosed.area is enclosed is constant..but still results are same.why this happening??

Acc to Faraday's law, induced emf is numerically equal to the rate of change of flux associated with a loop. If there is a conductor (no loop) then we can consider an imaginary loop by taking fix wires including this conductor of which area increases due to motion in conductor (area swept by it). In this case loop is not closed so no current will flow but same emf is only induced in moving conductor not in fixed imaginary wire. To follow lectures in sequence use site physicsgalaxy com.

thank you very much sir.

Oh my God, your voice.

Why is P high potential and Q low potential?

sorry i think there is a mistake in the first calculation. You are missing are minus sign in the induced emf

SIR IF THERE IS A ROTATING AND TRANSLATING ROD IN FRONY OF AN INFIINITE LONG CURRENT CARRING WIREHW TO FIND EMF —-IS IT LIKE THIS LET THE LISTANCE FRM IT BE R AND TREATING IT AS CONSTANT VELOCITY PERPENDICULR TO WIRE IS Vcm+XW .. EMF=INTEGRATION XWPROJECTION IN DIRECTION OF VCM+Vcm x DX xB

SIR I AM TELLING IF THERE IS A ROD WHICH IS ROTATING AS WELL AS TRANS LATINGFRM A DISTANCE SAY X OF INFINUTE LONG CURRENT CARRING CONDUCTOR ITS EMF CLCULATION

+physics galaxy on account of your comment on 26 Dec 2012,what will happen if a closed loop is moving(or rotating) in a uniform magnetic feild perpendicular to it..will a emf will induced or not ?? please explain both the cases i.e. moving(or rotating)

besttttttttttt video on this topic for iit

This emf does not depend on the angle covered by the rod. Why?

Thank you sir

Dear sir ,, if a rectangular loop is taken instead of rod with its plane perpendicular to the field and is rotating with angular speed omega then what will be it's induced EMF ans given is zero but acc. To this concept it shouldn't be zero as it is cutting mag field lines sir plz help..

Finally I got a good and simple to understand Video …

Can u plzzz explain what will happen to EMF if angular velocity (omega) is not constant ???

Sir how can I find the direction of high potential in first slide because motion is in circular path.

Which will be high potential side when rod is moved along axis passing through its COM?

sir how u calculated the area?

please help

dS=half πR^2 dtheta

You have forgotten π in calculating dS

sir in case of rotation of wheel why we not use no of spokes to calculate induced emf

sir if instead of a conducting rod, a conducting disc is used the answers remains same. How is that possible sir bcoz the magnetic flux is not at all changing

Sir if a disc is rotating with some angular velocity in a uniform magnetic field then emf would be induced or not? You have answered this type of query in a previous comment but I didn't get across which points emf would be induced? And sir what about two diametrically opposite points ? Will there be a potential difference between them?

it is faraday's law not farraeyde law

EDIT:- XD

Sir how did u write the value of ds

My question is that rod has shifted from initial position to new position but here, there is no surface which is closed through which flux can be changed then how emf can be induced

Is cutting flux is only sufficient condition for inducing emf?

is in this case also in steady state current is zero

Sir why the radius become half in faraday law

What if the rod were stationary and the field was changing at dB/dt?

Sir so at steady steady net force towards the centre will be = centripetal (non inertial frame) force (for an element) so qvb-qe=mv2/r ! Is it true ?

I didnt get that how emf is produced in this ? In this case area is not changing , anlgle of magnetic field and area vector is not changing , magnetic field is also constant then how flux will change ? Please explain

even magnetic fluxwill besame on integrating bwlsqquare/2 ?i got