Find the motional emf in a closed loop ABCD. Answer comes out to be 2RBV.
I have two doubts In it's concept.
1) here why we are calculating the emf by breaking the loop into two halves instead of considering effective length to be 0 since it is a closed loop?
2) in the 2nd figure attached below question figure we are saying that emf induced in 2nd loop is 0 since flux through it is constant. Why we are not considering this case in this question.
I request you to answer my these 2 doubts in detail, so that my concept becomes clear
Asked by rohitraman1115 | 15th Mar, 2021, 08:29: PM
when a loop made of conducting wire is moved in a region of magnetic field , we get induced EMF E in the loop
according to Faraday's law of induction as described by the following equation.
E = -dΦ/dt .........................(1)
Where Φ is flux of magnetic field passing through area A of loop so that Φ = ( B × A ) if B is flux density of magnetic field.
Equation (1) states that magnitude of induced EMF is directly proportional to rate of change of flux of magnetic field
passing through area of loop.
If flux of magnetic field increases , then induced EMF is negative .
If flux of magnetic field decreases , then induced EMF is positve .
If a circular loop of conducting wire enetrs or exits in a uniform magnetic field region ,
then induced EMF in the loop is determined as follows
Hence in uniform magnetic field , induced EMF is directly proportional to rate of change of area that
overlaps with flux of magnetic field.
For a circular loop , calculation of rate of change of area overlapping flux of magnetic field is very difficult.
Since induced EMF depends only on area and is indepenedent of length of wire , circular loop is equated to
a square loop that has same area of circular loop . The side of equivalent square loop is calculated as follows
a2 = π R2 or a = ( √π ) R
The process of getting induced EMF, when a square loop is moved in uniform magnetic field is described below
When square loop enters in magnetic field region , as shown in above figure , part of area of loop that
encloses the flux of magnetic field increases. Hence as per eqn.(2) , negative EMF is induced in square loop .
Similarly, when square loop exits out of magnetic field region , part of area of loop that encloses the flux of
magnetic field decreases. Hence at the time of exit, positive EMF is induced in square loop.
When the square loop is fully inside the magnetic field, there is no change in magnetic flux and hence induced EMF is zero.
Magnitude of induced EMF is calculated as follows
we get from eqn. (2) , E = B × (dA/dt) = B × a × da/dt = B × a × v
where v is speed of loop movement .
Regarding your doubts
(1) The loop is not broken into two halves to get induced EMF. As shown in figure , when loop enters
magnetic field region, flux density of magnetic field increases with respect to time and that induces negative EMF .
Similarly when the loop exits magnetic field region , flux density decreases and we get positive EMF .
Length of loop does not play any role to get induced EMF only area of loop matters.
(2) When the loop is completely inside magnetic field region, movement of loop does not create change in magnetic flux.
Hence, if there is no change in magnetic flux , then no induced EMF .
This is very much considered while analysing the induced EMF due to movement of loop made of conducting wire
in magnetic field. Graph of induced EMF shown in figure indicates that induced EMF is zero, when the loop is moving
inside uniform magnetic field region.
Answered by Thiyagarajan K | 16th Mar, 2021, 11:31: AM
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