Self Studies

Electromagnetic Induction Test - 57

Result Self Studies

Electromagnetic Induction Test - 57
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    A thin copper wire of length 100 meters is wound as a solenoid of length λ\lambda and radius r. Its self-inductance is found to be L. Now if the same length of wire is wound as a solenoid of length λ\lambda but the radius r/2, then its self inductance will be:
    Solution
    As we know that,   self inductance = μ  n2lA{ \mu  }_{ \circ  }{ n }^{ 2 }lA
     where n= number of turns per unit length
                 l=length of solenoid
                 A=cross section
    So, self inductance(L) =μ  n2lπR2{ \mu  }_{ \circ  }{ n }^{ 2 }l\pi { R }^{ 2 }
                                  L=μ  (Nl ) 2lπR2\Rightarrow L={ \mu  }_{ \circ  }{ \left( \frac { N }{ l }  \right)  }^{ 2 }l\pi { R }^{ 2 }                                                 (Where N= Total number of turns)
                                  L=μ  N2πR2l\Rightarrow L=\frac { { \mu  }_{ \circ  }N^{ 2 }\pi { R }^{ 2 } }{ l }
    Now, it is given that when length is λ\lambda and radius is'r' then ,
                    L=μ o(1002πr ) 2πr2λ  { { L } }={ \mu  }_{ o }{ \left( \frac { 100 }{ 2\pi r }  \right)  }^{ 2 }\frac { \pi { r }^{ 2 } }{ \lambda  } 
    Now, in second case length is λ\lambda and radius is 'r/2' then,
                   L=μ o(1002πr2  ) 2π λ (r2 ) 2{ { L } }'={ \mu  }_{ o }{ \left( \frac { 100 }{ 2\pi \frac { r }{ 2 }  }  \right)  }^{ 2 }\frac { \pi  }{ \lambda  } { \left( \frac { r }{ 2 }  \right)  }^{ 2 }
             L=μ o(1002πr ) 2πr2λ =L\Rightarrow{ { L } }'={ \mu  }_{ o }{ \left( \frac { 100 }{ 2\pi r }  \right)  }^{ 2 }\frac { \pi { r }^{ 2 } }{ \lambda  } =L
             L=L\Rightarrow L'=L

  • Question 2
    1 / -0
    Two coils of self inductance 6mH6 mH and 8mH8 mH are connected in series and are adjusted for highest coefficient of coupling. Equivalent self inductance LL for the assenbly is approximately 
    Solution

  • Question 3
    1 / -0
    A coil of inductance 300 mH and resistance 2Ω2 \Omega  is connected to a source of voltage 2V. The current reaches half of its steady state value in
    Solution
    Hence, option BB is the correct answer.

  • Question 4
    1 / -0
    The inductance between AA and DD is:

    Solution

  • Question 5
    1 / -0
    The dimensions of self- induction are :

    Solution
    ϕ =LI\phi  =LI

    and also ϕ =B.A=B2\phi  =B.A = B^{2}

    [B]MLT2LI1AT\frac{MLT^{2}}{LI^{-1}A T}

    [B]MLT2LA [B]\frac{MLT^{-2}}{LA}

    [B]MT2A1 [B]MT^{-2}A^{-1}

    [ϕ ]=MT2L2A1 [\phi  ]=MT^{-2}L^{2}A^{-1}

    thee finally dimension of inductance
    [L]=[ϕ ][I]\frac{[\phi  ]}{[I]}
    [L]=MT2C2A2 MT^{-2}C^{2}A^{-2}

  • Question 6
    1 / -0
    A conducting ring of radius r is rolling without slipping with a constant angular velocity ω \omega ( given figure). if the magnetic field strength is B and is directed into the page then the emf induced across PQ is:

    Solution
    Solution:\textbf{Solution}:
    emf induced is given by,
    E=Bl22ω\left | \vec{E} \right |=\dfrac{Bl^2}{2}\omega
    where l=2r l = \sqrt2 r
    \therefore E=B×(2r)22ω\left | \vec{E} \right |= \dfrac{B\times (\sqrt 2 r)^2}{2}\omega = Bωr2B\omega r^2 

    Hence A is the correct option\textbf{Hence A is the correct option}
  • Question 7
    1 / -0
    A conducting wire frame is placed in a magnetic field which is directed in to the paper. The magnetic field increasing at a constant rate. The directions of induced current in wires AB and CD are:

    Solution
    Solution:Solution: Inward magnetic field (')
    increasing. Therefore, induced
    current in both the loops should be
    anticlockwise. But as the area of
    loop on right side is more, induced
    emf in this will be more compared
    to the left side loop

    (e=dφdt=A.dBdt)\left(e=-\frac{d \varphi}{d t}=-A . \frac{d B}{d t}\right)

    Therefore net current in the
    complete loop will be in a direction
    shown above.
    So,theSo,the correctcorrect option:Aoption:A

  • Question 8
    1 / -0
    The inductance of a solenoid 0.5 m0.5\ m long of cross-sectional area 20 cm220\ cm^2 and with 500500 turns is
    Solution

  • Question 9
    1 / -0
    The current in a coil decreases from  1A1 A  to  0.2A.0.2 A .  In  1010  sec. Calculate the coefficient of self-inductance. If induced emf is  0.40.4  volt.
    Solution
    e=Ldidte = -L \dfrac{di}{dt}
    0.4=L(0.21)10\Rightarrow 0.4 = -\dfrac{L(0.2 - 1)}{10}
    L=5H \Rightarrow L = 5 \,H
  • Question 10
    1 / -0
    If the magnetic flux linked up with a coil is changed by 40%, the energy storied in the coil changes by 
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now