Self Studies

Electromagnetic Induction Test - 25

Result Self Studies

Electromagnetic Induction Test - 25
  • 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
    In an AC generator, a coil with $$N$$ turns, all of the same area $$A$$ and total resistance $$R$$, rotates with frequency $$1 rad/s$$ in a magnetic field $$B$$. The maximum value of emf generated in the coil is:

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

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

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

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

    now if, $$\sin\omega t=1$$ (maximum),
    then             $$e_{max}=NBA\omega $$,

    if $$\omega=1rad/s$$,
    then             $$e_{max}=NBA$$
  • 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 $$72 km/h$$ along the track. The reading of the millivoltmeter (in $$mV$$ ) is :

    (Vertical component of the earths magnetic induction is $$2\times 10^{-5}T$$ )
    Solution
    $$Emf= Bvl$$
    $$= 2\times 10^{-5}\times 72\times \dfrac{1000}{3600} \times 1$$
    $$ = 4 \times 10^{-4}V$$
     $$= 0.4mV$$ 
  • Question 4
    1 / -0
    The area of a coil is $$500cm^{2}$$ and the number of turns in it is $$2000$$. It is kept perpendicular to a magnetic field of induction $$4 \times10^{-5}Wb /m^{2}$$. The coil is rotated through $$180$$ in $$0.1$$ second. If the resistance of the total circuit is $$20\Omega$$, then the value of the induced charge flowing in the circuit will be :
    Solution
    Answer is B
    Initial flux through the coil, $$= BA \cos \theta$$
    $$= BA \cos 0^o$$, as the coil is perpendicular to magnetic field
    $$= BA  \ \ Wb$$

    Final flux after the rotation,

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

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

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

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

    From (i) & (ii)

    $$Q=\dfrac{2NBA}{R} = \dfrac{2\times2000\times4\times10^{-5}\times500\times10^{-4}}{20} = 4\times10^{-4}C$$ 
  • 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 $$L\propto N^{2}$$ .
    Solution
    Both are true $$L\  \alpha \  N^2$$
  • Question 6
    1 / -0
    A current of $$2A$$ is passed through a coil of $$1000$$ turns to produce a flux of $$0.5\mu  Wb$$. Self inductance of the coil 
    Solution
    $$N\phi =Li$$
       $$L=\dfrac { N\phi  }{ L } $$
           = $$\dfrac { 1000\times 0.5\times { 10 }^{ -6 } }{ 2 } $$
            $$= 2.5 \times 10^{-4} H$$
  • Question 7
    1 / -0
    Two ends of a conducting rod of varying cross-section are maintained at $$200^\circ C$$ and $$0^\circ C$$ 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 $$L = 2 mH$$, current flow through it is I = $$t^2 e^{-t}$$ then the time at which emf become zero : -
    Solution
    $$I = t^2 e^{-t}$$

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

    $$\dfrac{dI}{dt} = 2te^{-t}  - t^2 e^{-t} = 0$$

    $$t = 2sec$$

    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 $$4 cm^2$$. Then the coefficient of self induction is approximately
    Solution
    $$n=400 turns, l=20 cm, A=4 cm^2$$
    $$L=?$$
    $$L=\frac {1\cdot 25\times 10^{-6}Hm^{-1}\times 400\times 400\times 4\times 100}{100\times 100\times 20}$$
    $$=4\times 10^{-4}H$$
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