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Electromagnetic Induction Test - 58

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Electromagnetic Induction Test - 58
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  • Question 1
    1 / -0
    A wire of fixed length is wound on a solenoid of length l and radius r. Its self inductance is found to be L. Now, if same wire is wound on a solenoid of length $$\dfrac{l}{2}$$ and radius $$\dfrac{r}{2}$$, then self inductance will be
    Solution
    $$L=\mu _{0}\times n^{2}\times V$$    (n $$\rightarrow $$ Number of turns per unit length , V $$\rightarrow $$ Volume of cylinder)
    $$=\mu _{0}\times \frac{N^{2}}{l^{2}}\times \pi r^{2}\times l$$
    $$L=\frac{\mu _{0}N^{2}\pi r^{2}}{l}$$
    Let length of wire be $$l_{1}$$
    $$l_{1}=(2\pi r)\times N \Rightarrow N=(\frac{l_{1}}{2\pi r})$$
    $$L=\frac{\mu _{0}\pi r^{2}}{l}\times \frac{l_{1}^{2}}{4\pi ^{2}r^{2}}=\frac{\mu _{0}l_{1}^{2}}{4\pi l}$$
    $$\therefore  L \propto \frac{1}{l}\Rightarrow \frac{L_{2}}{L_{1}}=\frac{l_{1}}{l_{2}}=\frac{l}{\frac{l}{2}}=2$$
    $$L_{2}=2L_{1}$$

  • Question 2
    1 / -0
    Electro motive force induced by motion of conductor across magnetic field is called
    Solution

  • Question 3
    1 / -0
    A cylindrical magnet is kept along the axis of a circular coil. On rotating the magnet about its axis. the coil will have induced in it
    Solution
    No, Current will not be induced in the coil if the coil is rotated about it's axis
    Formula for the flux is.      $$\phi$$ = NBA = constant
    $$\therefore$$                                            i = $$\frac{d\phi}{dt}$$ = 0
                                                   i = 0


  • Question 4
    1 / -0
    Magnetic flux in the circuit containing a coil of resistance 2 ohm changes from 2.0 wb to 10 wb in 0.2 sec. the charge passed through the coil in this time is 
  • Question 5
    1 / -0
    A magnetic field of flux density $$10 T$$ acts normal to a coil of $$50$$ turns having $$100$$ $$cm^2$$  area. The e.m.f. induced if the coil is removed from the magnetic field in $$0.1$$ seconds is:
    Solution
    In this case, $$B=10T\, ,N=50\,turns\, ,A=100\,cm^2=10^{-2}m^2$$ and $$dt=0.1\,s$$

    We know that,

    $$\phi_1=NBA\,Cos\,\theta=NBA$$ (as $$Cos\theta=Cos0=1$$)

    Therefore,

    $$\phi_1=50\times 10\times 10^{-2}=5\,weber$$

    When the coil is removed from the magnetic field, $$B=0$$

    So,

    $$\phi_2=0$$

    Therefore,

    $$\phi_1-\phi_2=d\phi=0-5=-5\,weber$$.

    We also know that $$e=-\dfrac{d\phi}{dt}=\dfrac{-5}{0.1}=50\,V$$

    Hence, the induced emf if the coil is removed from the magnetic field in  $$0.1$$ second is $$50\,V$$.
  • Question 6
    1 / -0
    A constant current $$I_{0}$$ is passing through a long straight wire (shown in figure).
    A rectangular loop of total resistance $$R$$ is moving parallel to the wire. Then:

    Solution
    The magnetic flux in the loop remains constant induced emf in the loop is zero.
    Hence, induced current in loop is zero.
  • Question 7
    1 / -0
    When current changes from $$13\ A$$ to $$7\ A$$ in $$0.5\ sec$$ through a coil, the emf induced is $$3\times 10^{-4}\ V$$. The coefficient of self induction is:
    Solution

  • Question 8
    1 / -0
    The magnetic flux $$\phi$$ (in weber) in a closed circuit of resistance $$10\ ohm$$ varies with time $$t$$ (in second) according to equation $$\phi=6t^{2}-5t+1$$. The magnitude of induced current at $$t=0.25\ s$$ is:
    Solution

  • Question 9
    1 / -0
    Self inductance of coil measures
    Solution

  • Question 10
    1 / -0
    Which type of current can be obtained from an electric generator?
    Solution

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