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Thread: Question of Electronic question

  1. #1
    Fuchcha FaaDoO Engineer
    Join Date
    Apr 2011

    Gender: : Female

    City : Pune

    Question of Electronic question

    A conductor in (x-y) plane and having lenth 1m is moving with velocity
    V=(2i+3j+k)m/sec.Magnetic flux desity B=(i+2j)wb/m2.Potential difference at the
    end of the conductor is(Explain your answer)

    i)0 ii)4.88V iii)root6 iv)none

    2)Two parallel plate separated by x are connected by wire.point charge q is located between
    platesat a distance x1 from one plate and x-x1 fromm onother plate.
    surface charge on the inner surface of each plate is(explain your answer)

    i)0 ii)q/2 iii)q iv)-q

    3)A 100W lamp radiates all of its energy in the form of E=E0 sin(2P(vt-x)/l) ;H=H0 sin(2P (vt-x)/l)
    The energy emitted uniformly in all direction.The max. electric and magnetic intensity at 200cm
    from lamp(explain your answer)

    i)380.6V/m,0 ii)38.6V/m,.103A/m iii)0,.103A/m iv)0, 0

    4)Variying magnetic field at the rate of dB/dt within a cylindrical region of radius r.This produce
    electric field E at any point at the loop with radius x(Explain your answer)

    i)E=-(db/dt)*x for x<r ii)E=-(dB/dt).x for x>r iii)E=(dB/dt)*(x/2) for x<r iv)E=(dB/dt)*x/2
    for x/2 <r v)E=(dB/dt)*r^2/(2*x) for x>r.
    Ans:4 and 5

    5)four charges in corner of the square of side root2.The E at P at distance R is given by
    (Explain your answer)

    6)The proton released from rest at a distance of 10^-10m from another proton(Explain your answer)
    Ans:The velocity of the released proton would go on increasing and finally equal to
    50Km/sec ,at infinite distance apart.

    7)75ohm transmission line is terminated by two loads R1 and R2 such that
    both have same SWR.The value of R1 and R2.(Explain your answer)

    i) 250,200 ii)225,25 iii)100,150 iv)50,125

  2. #2
    Fuchcha FaaDoO Engineer
    Join Date
    Apr 2011

    Gender: : Male

    City : Porbandar

    (1) As u might know... F=q(vXB)
    then F/q=E=Vd=vXB
    now d=1 so u can have ans as V=magnitude of vXB as root6

    (2) As per Gauss law... Take two plate as closed Gaussian surface then we have enclosed charge as q. now two plate have total induced charge as q so one have half q/2.

    (5)Find E and v by superposition. and ans is Ans:9q(2sin^2-1)/(4*pi*Effsalan*R^4)

    (6)As two proton tend to repel each other, When they are freed, tehy produce repulsive force on each other. Thus their potential energy is converted into the kinetic energy. Find v in 1/2*m*v^2=k*q^2/r^2

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