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Electromagnetic Induction Test - 25

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Electromagnetic Induction Test - 25
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  • Question 1
    1 / -0
    In an AC generator, a coil with NN turns, all of the same area AA and total resistance RR, rotates with frequency 1rad/s1 rad/s in a magnetic field BB. The maximum value of emf generated in the coil is:

    Solution
    Let at time t=0t=0, the coil is vertical  and at time tt , plane of coil makes an angle θ\theta with the vertical, then 
                       θ=ωt\theta=\omega t,   (ω=\omega= uniform angular velocity),

    in this position, magnetic flux linked with coil will be,
                       ϕ=NBAcosθ\phi=NBA\cos \theta,   (where, A=A= area of coil,)
    or               ϕ=NBAcosωt\phi=NBA \cos \omega t,

    now, differentiating this equation w.r.t. time, we get
                       dϕdt=ddt(NBAcosωt)\dfrac{d\phi}{dt}=\dfrac{d}{dt}(NBA \cos\omega t),
    or                dϕdt=NBAωsinωt\dfrac{d\phi}{dt}=-NBA\omega\sin\omega t,

    if $e$$ is the emf induced in coil then by Faraday's law,
                       e=dϕdt=NBAωsinωte=-\dfrac{d\phi}{dt}=NBA\omega\sin\omega t,

    now if, sinωt=1\sin\omega t=1 (maximum),
    then             emax=NBAωe_{max}=NBA\omega ,

    if ω=1rad/s\omega=1rad/s,
    then             emax=NBAe_{max}=NBA
  • Question 2
    1 / -0
    Assertion : When two coils are wound on each other, the mutual induction between the coils is maximum.
    Reason : Mutual induction does not depend on the orientation of the coils.
    Solution
    Mutual inductance is purely a geometric quantity, depending only on size, no. of turns, relative position and relative orientation of the 2 coils .
  • Question 3
    1 / -0
    Two parallel rails of a railway track insulated from each other and with the ground are connected to a millivoltmeter. The distance between the rails is one metre. A train is traveling with a velocity of 72km/h72 km/h along the track. The reading of the millivoltmeter (in mVmV ) is :

    (Vertical component of the earths magnetic induction is 2×105T2\times 10^{-5}T )
    Solution
    Emf=BvlEmf= Bvl
    =2×105×72×10003600×1= 2\times 10^{-5}\times 72\times \dfrac{1000}{3600} \times 1
    =4×104V = 4 \times 10^{-4}V
     =0.4mV= 0.4mV 
  • Question 4
    1 / -0
    The area of a coil is 500cm2500cm^{2} and the number of turns in it is 20002000. It is kept perpendicular to a magnetic field of induction 4×105Wb/m24 \times10^{-5}Wb /m^{2}. The coil is rotated through 180180 in 0.10.1 second. If the resistance of the total circuit is 20Ω20\Omega, then the value of the induced charge flowing in the circuit will be :
    Solution
    Answer is B
    Initial flux through the coil, =BAcosθ= BA \cos \theta
    =BAcos0o= BA \cos 0^o, as the coil is perpendicular to magnetic field
    =BA   Wb= BA  \ \ Wb

    Final flux after the rotation,

    f=BAcos180of = BA \cos 180^o, the coil is rotated through 180o180^o
    =BA   Wb= -BA  \ \ Wb

    Therefore, estimated value of the induced emf (EE) as per Faraday's law is,

     E=Nt=N(BA(BA))t=2NBAt   E = \dfrac{N}{t} = \dfrac{N (BA-(-BA))}{ t} = \dfrac{2NBA}{ t}                    -------(i)

    Further, E=IR=QRtE= IR= \dfrac{QR}{ t}  -------(ii)

    From (i) & (ii)

    Q=2NBAR=2×2000×4×105×500×10420=4×104CQ=\dfrac{2NBA}{R} = \dfrac{2\times2000\times4\times10^{-5}\times500\times10^{-4}}{20} = 4\times10^{-4}C 
  • Question 5
    1 / -0
    Assertion : When number of turns in a coil is doubled, coefficient of self inductance of the coil becomes 4 times.
    Reason : This is because LN2L\propto N^{2} .
    Solution
    Both are true L  α  N2L\  \alpha \  N^2
  • Question 6
    1 / -0
    A current of 2A2A is passed through a coil of 10001000 turns to produce a flux of 0.5μ Wb0.5\mu  Wb. Self inductance of the coil 
    Solution
    Nϕ=LiN\phi =Li
       L=Nϕ LL=\dfrac { N\phi  }{ L }
           = 1000×0.5×1062\dfrac { 1000\times 0.5\times { 10 }^{ -6 } }{ 2 }
            =2.5×104H= 2.5 \times 10^{-4} H
  • Question 7
    1 / -0
    Two ends of a conducting rod of varying cross-section are maintained at 200C200^\circ C and 0C0^\circ C respectively, figure. In steady state

    Solution

  • Question 8
    1 / -0
    In the arrangement shown in given Figure, there are two coils wound on a non-conducting cylindrical rod. Initially the key is not inserted. Then the key is inserted and later removed. Then,

    Solution

  • Question 9
    1 / -0
    For a coil having L=2mHL = 2 mH, current flow through it is I = t2 ett^2 e^{-t} then the time at which emf become zero : -
    Solution
    I=t2etI = t^2 e^{-t}

    e=LdIdte = \dfrac{L dI}{dt}

    dIdt=2tet t2et=0\dfrac{dI}{dt} = 2te^{-t}  - t^2 e^{-t} = 0

    t=2sect = 2sec

    Hence (A) is the correct answer
  • Question 10
    1 / -0
    A solenoid (air core) ahs 400 turns, is 20 cm long and has a cross section of 4cm24 cm^2. Then the coefficient of self induction is approximately
    Solution
    n=400turns,l=20cm,A=4cm2n=400 turns, l=20 cm, A=4 cm^2
    L=?L=?
    L=125×106Hm1×400×400×4×100100×100×20L=\frac {1\cdot 25\times 10^{-6}Hm^{-1}\times 400\times 400\times 4\times 100}{100\times 100\times 20}
    =4×104H=4\times 10^{-4}H
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