Proof of Faraday's Law
It is evident from Faraday's experiments that whenever there is a change in the magnetic flux passing through a closed circuit, an electric current is induced in the circuit. It can be explained on the basis of Lorentz force.
Suppose a conducting-rod JK (Fig. 7) is being moved without friction on the arms of a U-shaped stationary conductor MNOL with a velocity v towards right. The conductor is situated in a magnetic field B perpendicular to the plane of paper directed downwards. Due to the motion of the conducting-rod, the free electrons present in the rod are acted upon by a magnetic (Lorentz) force Fm of magnitude qvB which takes the electrons from the end J of the rod to the end K. Since a closed circuit is available to the electrons, they drift along the path J-*K-*O^N-*J. Thus, an electric current is established in the circuit along J->N->0->K-*J (anticlockwise)*. So long the rod JK is kept moving in the magnetic field, the electric current continues to flow in the circuit. It means that an emf is induced in the moving rod which maintains the current in the circuit. Since this emf induced in the rod is due to the motion of the rod, it is also called as the 'motional emf. It is due to the Lorentz forces acting on the free electrons in the moving rod.
Suppose, the induced emf in the moving rod JK is e and the induced current in the circuit is i. We know that when a current-carrying conductor is in a magnetic field, it is acted upon by a force in a direction given by the Fleming's left-hand rule. The magnitude of the force imposed by the magnetic field B on the current-carrying rod JK is given by
F' = i IB,
where I is the length of the rod JK. The force F*' is directed towards left. Hence, to keep the rod moving with a constant velocity v towards right, a force F equal and opposite to F' will have to be applied on it; that is
F = -F' = -ilB.
Suppose, under the force Ft the rod JK undergoes a displacement Ax* in the direction of the force in a time-interval At and comes in the position J 'K'. Then, the work done on the rod will be given by
W =F*.Ax*= F Ax = -HB Ax. But A* = v x Af (magnitude of velocity x time-interval).
W = -HBvAt.
But i x At = q (charge flown through the circuit in time At)
W = -Bvlq.
This work provides the necessary energy for the flow of charge in the circuit. We know that the energy supplied by a cell in flowing unit charge through a circuit is called the emf of the cell. Hence the induced emf e in the rod / K (which is working as a cell in the circuit J N O K J) is given by
e = W/q = - Bv I. ...(i)
In the time-interval At, the area of the circuit increases from JNOKJ to J' N O K' JHence, during this time-interval, the change in the magnetic flux passing through the circuit is given by
A<Pb = magnitude of magnetic field x change in area (JJ 'K'K) = B x (I x Ax).
Hence the rate of change of magnetic flux is
AOfl Ax
I — O I ~~ .
At At
But Ax/At = v (velocity of the rod). Thus
-tf'Blv. ...(ii)
Comparing eq. (i) and (ii), we get
A<Dfl
e = -~AT■
In the limit At —» 0,
d<bB
e = ~~dT■
This is Faraday law of electromagnetic induction.
The negative sign signifies Lenz's law. The direction of current (anticlockwise) in the circuit due to the induced emf e is such that the force imposed on the rod (due to the current) opposes the motion of the rod (the motion of the rod is the cause of the current).
It can be seen in the other way also. The direction of current in the circuit is such that the magnetic field produced due to this current is just opposite to the original field B. That is, it opposes the increase in the magnetic flux passing through the circuit (the increase in the magnetic flux is the cause of the generation of current). We may see it from any point of view, the direction of the induced current is always such that it opposes the very cause of its production. This is Lenz's law.
Dimensions of Induced E.M.F.: Numerically, we have
d<bB e = ~dT-
, dimensions of e = = ^T^A'*] = a , A_,
dimensions of t [T]
It is evident from Faraday's experiments that whenever there is a change in the magnetic flux passing through a closed circuit, an electric current is induced in the circuit. It can be explained on the basis of Lorentz force.
Suppose a conducting-rod JK (Fig. 7) is being moved without friction on the arms of a U-shaped stationary conductor MNOL with a velocity v towards right. The conductor is situated in a magnetic field B perpendicular to the plane of paper directed downwards. Due to the motion of the conducting-rod, the free electrons present in the rod are acted upon by a magnetic (Lorentz) force Fm of magnitude qvB which takes the electrons from the end J of the rod to the end K. Since a closed circuit is available to the electrons, they drift along the path J-*K-*O^N-*J. Thus, an electric current is established in the circuit along J->N->0->K-*J (anticlockwise)*. So long the rod JK is kept moving in the magnetic field, the electric current continues to flow in the circuit. It means that an emf is induced in the moving rod which maintains the current in the circuit. Since this emf induced in the rod is due to the motion of the rod, it is also called as the 'motional emf. It is due to the Lorentz forces acting on the free electrons in the moving rod.
Suppose, the induced emf in the moving rod JK is e and the induced current in the circuit is i. We know that when a current-carrying conductor is in a magnetic field, it is acted upon by a force in a direction given by the Fleming's left-hand rule. The magnitude of the force imposed by the magnetic field B on the current-carrying rod JK is given by
F' = i IB,
where I is the length of the rod JK. The force F*' is directed towards left. Hence, to keep the rod moving with a constant velocity v towards right, a force F equal and opposite to F' will have to be applied on it; that is
F = -F' = -ilB.
Suppose, under the force Ft the rod JK undergoes a displacement Ax* in the direction of the force in a time-interval At and comes in the position J 'K'. Then, the work done on the rod will be given by
W =F*.Ax*= F Ax = -HB Ax. But A* = v x Af (magnitude of velocity x time-interval).
W = -HBvAt.
But i x At = q (charge flown through the circuit in time At)
W = -Bvlq.
This work provides the necessary energy for the flow of charge in the circuit. We know that the energy supplied by a cell in flowing unit charge through a circuit is called the emf of the cell. Hence the induced emf e in the rod / K (which is working as a cell in the circuit J N O K J) is given by
e = W/q = - Bv I. ...(i)
In the time-interval At, the area of the circuit increases from JNOKJ to J' N O K' JHence, during this time-interval, the change in the magnetic flux passing through the circuit is given by
A<Pb = magnitude of magnetic field x change in area (JJ 'K'K) = B x (I x Ax).
Hence the rate of change of magnetic flux is
AOfl Ax
I — O I ~~ .
At At
But Ax/At = v (velocity of the rod). Thus
-tf'Blv. ...(ii)
Comparing eq. (i) and (ii), we get
A<Dfl
e = -~AT■
In the limit At —» 0,
d<bB
e = ~~dT■
This is Faraday law of electromagnetic induction.
The negative sign signifies Lenz's law. The direction of current (anticlockwise) in the circuit due to the induced emf e is such that the force imposed on the rod (due to the current) opposes the motion of the rod (the motion of the rod is the cause of the current).
It can be seen in the other way also. The direction of current in the circuit is such that the magnetic field produced due to this current is just opposite to the original field B. That is, it opposes the increase in the magnetic flux passing through the circuit (the increase in the magnetic flux is the cause of the generation of current). We may see it from any point of view, the direction of the induced current is always such that it opposes the very cause of its production. This is Lenz's law.
Dimensions of Induced E.M.F.: Numerically, we have
d<bB e = ~dT-
, dimensions of e = = ^T^A'*] = a , A_,
dimensions of t [T]
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