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Electromagnetic Induction Test - 57

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Electromagnetic Induction Test - 57
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  • Question 1
    1 / -0
    A thin copper wire of length 100 meters is wound as a solenoid of length $$\lambda$$ 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 $$\lambda$$ but the radius r/2, then its self inductance will be:
    Solution
    As we know that,   self inductance = $${ \mu  }_{ \circ  }{ n }^{ 2 }lA$$
     where n= number of turns per unit length
                 l=length of solenoid
                 A=cross section
    So, self inductance(L) =$${ \mu  }_{ \circ  }{ n }^{ 2 }l\pi { R }^{ 2 }$$
                                  $$\Rightarrow L={ \mu  }_{ \circ  }{ \left( \frac { N }{ l }  \right)  }^{ 2 }l\pi { R }^{ 2 }$$                                                 (Where N= Total number of turns)
                                  $$\Rightarrow L=\frac { { \mu  }_{ \circ  }N^{ 2 }\pi { R }^{ 2 } }{ l } $$
    Now, it is given that when length is $$\lambda$$ and radius is'r' then ,
                    $${ { L } }={ \mu  }_{ o }{ \left( \frac { 100 }{ 2\pi r }  \right)  }^{ 2 }\frac { \pi { r }^{ 2 } }{ \lambda  } $$
    Now, in second case length is $$\lambda$$ and radius is 'r/2' then,
                   $${ { L } }'={ \mu  }_{ o }{ \left( \frac { 100 }{ 2\pi \frac { r }{ 2 }  }  \right)  }^{ 2 }\frac { \pi  }{ \lambda  } { \left( \frac { r }{ 2 }  \right)  }^{ 2 }$$
             $$\Rightarrow{ { L } }'={ \mu  }_{ o }{ \left( \frac { 100 }{ 2\pi r }  \right)  }^{ 2 }\frac { \pi { r }^{ 2 } }{ \lambda  } =L$$
             $$\Rightarrow L'=L$$

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

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

  • Question 4
    1 / -0
    The inductance between $$A$$ and $$D$$ is:

    Solution

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

    Solution
    $$\phi  =LI$$

    and also $$\phi  =B.A = B^{2}$$

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

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

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

    $$ [\phi  ]=MT^{-2}L^{2}A^{-1}$$

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

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

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

    $$\textbf{Hence A is the correct option}$$
  • 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
    $$Solution:$$ 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

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

    Therefore net current in the
    complete loop will be in a direction
    shown above.
    $$So,the$$ $$correct$$ $$option:A$$

  • Question 8
    1 / -0
    The inductance of a solenoid $$0.5\ m$$ long of cross-sectional area $$20\ cm^2$$ and with $$500$$ turns is
    Solution

  • Question 9
    1 / -0
    The current in a coil decreases from  $$1 A$$  to  $$0.2 A .$$  In  $$10$$  sec. Calculate the coefficient of self-inductance. If induced emf is  $$0.4$$  volt.
    Solution
    $$e = -L \dfrac{di}{dt} $$
    $$\Rightarrow 0.4 = -\dfrac{L(0.2 - 1)}{10}$$
    $$ \Rightarrow L = 5 \,H$$
  • 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 
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