241)ques is wrong,,,max emf induced when switch is closed
2)a,c
3)b,d
plz let me know if i am rite or rong..
Q(1) An inductor is connected to a battery through a switch. Explain why the emf induced in the inductor is much larger when the switch is opened as compared to the emf induced when switch is closed .( not a doubt)
Q(2) A metal sheet is placed in front of a strong magnetic pole. A force is needed to :
(A) hold the sheet there if the metal is magnetic
(B) hold the sheet there if the metal is non-magnetic
(C) move the sheet away from the pole with uniform velocity if metal is magnetic
(d) move the sheet away from the pole with uniform velocity if metal is nonmagnetic
(Neglect any effects of para, ferro, diamagnetism and gravity )
Q(3)An LR circuit with a battery is connected at t =0. Which of the following quantities is not zero just after the connection ?
(a) current in the circuit
(b)magnetic field in the inductor
(c) power delivered in the battery
(d)emf induced in the conductor
PS: all are more than one ans correct .2nd and 3rd q are my doubts
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12 Answers
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23 ur 2nd and 3rd ans are wrong .
for 2nd ques i thought the same ans i.e a and c but ans is a,c,d .[2]
btw Q1 isnt my doubt and hence i m sure abt ques as well as d ans [1]
24ya 2 should be a,c,d
and 1st.. induced emf is max at t=0..thats why we remove inductor at that point....and at t=∞...emf induced is zero...
1Answer to Qsn 2.
There is no effect of gravity, ferromagnetism, diamagnetism, etc. Hence no external forces act on the plate except the magnetic force.
In case a :
Hence the plate will be held by the magnetic force alone.
If they mean to ask that the plate has to be held at a distance from the magnet, then yes, a force is required.
Otherwise if they mean to ask that the plate has to stick to the magnet and stay there, then another force is not required.
In case b :
This is a serious joke... its obvious that a force is not required since magnetic force will not affect the plate.
In case c :
A force is required to move the sheet away since the plate is attracted by the magnet. A magnetic force is exerted on the plate, so a force is needed to counteract this force.
In case d :
The plate is non magnetic, so a force is required to start uniform motion, but once it moves with constant velocity, no force is required.
1Answer to Qsn.3
Consider all A/C terms as sine wave functions.
Current is a sine wave since the current is A/C. So it varies with time, and at t=0, it is 0.
Magnetic field inside the inductor will depend on current, so field is induced only when there is current. Hence field is 0 at t=0.
Emf is induced inside the inductor only when there is a change in current with time. Since the current is alternating, hence emf is given by
E = ∂$dt where $ is magnetic flux, given by B x A
Now, A is constant for inductor, and B depends on I. Since I is changing constantly, the rate of change of current is a constant. In other words,
E = A x ∂Bdt = A x u0 ∂Idt
So emf induced is a constant, not equal to 0. (Provided rate of change of current is constant)
Since Power is given by P = E x I where both E and I are sine functions, hence it is 0 at t=0.
Hence the correct answer in this case should be d
Now consider everything as cosine functions.
Then a,b are not 0. Power is still 0.
Hence the correct answers should be a,b,d
1I still didn't understand either of u guyz' solutions for the 1st qsn....
When you run current through an inductor, a magnetic field is built up through the coil. The magnetic field stores energy.
A changing magnetic field through an inductor induces a voltage in the inductor. Also, the faster the magnetic field changes, the larger is the induced voltage.
When the switch is first closed, the magnetic field is changing the fastest. It is built up in an (inverse) exponential fashion -- the field strength rises rapidly at first, and slows down.
So when the switch is first closed, the inductor's magnetic field is changing very rapidly, so the induced voltage across the coil is very large. As the circuit "settles," the magnetic field increases more slowly, so the induced voltage across the coil decreases.
After a period of time (which can range from microseconds to seconds, depending on the size of the inductor), the circuit is stable -- the field through the inductor is constant, which means there is zero voltage across it. In a steady-state condition, the inductor acts as a short circuit. It is sometimes said that inductors are shorts to DC.
23ankur , for second , no need of AC , simply think in terms of DC
btw ankur, tnx for the theory of magnetism ,but u dont need to write it ,its understood , ans is only d for 3rd
for Q1 i havent given any solution yet !!! ,
so for Q 1
\varepsilon =\frac{-d\phi }{dt}=-L\frac{di}{dt}\approx -L\frac{\Delta i}{\Delta t}
...i.e avg EMF
|\Delta i|=|i_{final} - i_{initial}|
let the time taken to open and close the switch be t
when u r closing the switch ,
i_{initial}=0
i_{final}=i_{o}(1-e^{\frac{-Rt}{L}})
hence i_{final}<i_{o}
but when opening the switch
i_{final}= 0
i_{initial}\approx i_{o}
hence
\Delta i_{opening} >\Delta _{i}_{closing}
hence EMF induced during opening > EMF induced while closing
( similar method is used by HCV in a solved example [1])
1Qwerty, if you are talking about Qsn 3 in terms of DC, thn all i can say is, the inductor acts as an open path without offering any resistance.
Then the current produced is instantaneous, and is not 0.
Magnetic field produced is not 0 since inductor acts like a solenoid.
Instantaneously induced emf is also not 0 since emf is induced immediately after t=0 due to change in current from 0 to constant value.
Power drawn from the battery is also not 0.
Btw i dont knw those formulae abt current n stuff tht u used in Q1... cn u explain me those?
23yaar these are related to induced EMF and Inductance , refer any standard book like HCV or resnick , u will get it :)