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Electromagnetic Induction Test - 68

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Electromagnetic Induction Test - 68
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  • Question 1
    1 / -0
    $$AB$$ is a resistanceless conducting rod which forms a diameter of a conducting ring of radius $$r$$ rotating in a uniform magnetic field $$B$$ as shown in figure. The resistors $${R}_{1}$$ and $${R}_{2}$$ do not rotate. Then the current through the resistor $${R}_{1}$$

    Solution
    Emf induced between A and center of circle or B and center of circle 

    $$=\int_0^rB(r\omega)dr=\dfrac{1}{2}B\omega r^2$$

    This is the also the potential difference across resistor $$R_1$$.

    Hence, $$\dfrac{1}{2}B\omega r^2=i(R_1)$$

    $$\implies i=\dfrac{B\omega r^2}{2R_1}$$
  • Question 2
    1 / -0
    A vertical conducting ring of radius $$R$$ falls vertically with a speed $$V$$ in a horizontal uniform magnetic field which is perpendicular to the plane of the ring. Which of the following statements is correct?

    Solution
    $$ EMF = \int _0 ^{\phi} Bvrd\theta (sin\theta)$$
    where $$\theta =0$$ at D
    Integrating
    $$V_{\theta}- V_D=  Bvr (1 -cos\theta)$$
    $$V_A = V_D+BvR$$
    $$V_C=V_D$$
    (Also can be seen by symmetry)
  • Question 3
    1 / -0
    A semicircular wire of radius $$R$$ is rotated with constant angular velocity about an axis passing through one end and perpendicular to the plane of the wire. There is a uniform magnetic field of strength $$B$$. The induced emf between the ends is

    Solution
    Induced motional emf in $$MNQ$$ is equivalent to the motional emf in an imaginary wire $$MQ$$.
    Therefore, 
    $${ e }_{ MNQ }={ e }_{ MQ }$$

    $$=\dfrac { 1 }{ 2 } { B\omega l }^{ 2 }$$

    $$={ 2B\omega R }^{ 2 }\quad \left(\because l=2R \right) $$

  • Question 4
    1 / -0

    Directions For Questions

    A standing wave $$y=2A \sin{kx} \cos{\omega t}$$ is set up in the wire $$AB$$ fixed at both ends by two vertical walls (see figure). The region between the walls contains magnetic field $$B$$. 

    ...view full instructions

    The wire is found to vibrate in the third harmonic. The maximum emf induced is

    Solution
    Since string vibrates in third harmonic, only a length of $$\lambda/3$$ will produce net emf.

    In a segment of length $$dx$$, $$dE=2A\omega sin(kx)dx$$.

    Integrating this in the range $$0$$ to $$\lambda/2$$, we get

    $$E=\dfrac { 4AB\omega  }{ k } $$.
  • Question 5
    1 / -0

    Directions For Questions

    A metal bar is moving with a velocity of $$v=5cm$$ $${s}^{-1}$$ over a U-shaped conductor. At $$t=0$$, the external magnetic field is $$0.1T$$ directed out of page and is increasing at a rate of $$0.2t$$ $${s}^{-1}$$. Take $$l=5cm$$, and at $$t=0,x=5cm$$

    ...view full instructions

    The emf induced in the circuit is

    Solution
    Area $$ A=lx$$

    Induced emf $$ { e=\dfrac { d\left( BA \right)  }{ dt }  }=lx\dfrac { dB }{ dt } +lB\dfrac { dx }{ dt } $$

    $$e={ lx\left( 0.2 \right)  }-{ lBv }\quad ...\left( x\quad is\quad decreasing \right) $$

    At t=0, $$e=0.05\times 0.05\times 0.2-0.05\times 0.1\times 0.05=250㎶$$
  • Question 6
    1 / -0
    A uniform thin rod of length L is moving in a uniform magnetic field $$\displaystyle B_{0}$$ such that velocity of its centre of mass is v and angular velocity is $$\displaystyle \omega=\frac{4v}{L}$$ Then

    Solution
    The charges inside the rod move under the magnetic force given by

    $$\vec{F}=q(\vec{v}\times \vec{B})$$

    Thus the charges experience force towards end P. Hence end P is at higher potential.

    EMF induced along a infinitesimal element along rod of length $$dx$$ is $$dV=Bv(dx)$$

    $$B(x\omega)dx$$

    Hence emf between points P and Q=$$\int_0^L B(x\omega )dx$$

    $$=\dfrac{1}{2}B\omega L^2$$

    $$=2BvL$$
  • Question 7
    1 / -0
    A thin semicircular conducting ring of radius $$R$$ is falling with its plane vertical in horizontal magnetic induction $$\vec { B } $$. At the position $$MNQ$$, the speed of the ring is $$V$$, and the potential difference developed across the ring is

    Solution
    $$E = \cfrac{d \phi}{dt}= B \cfrac{dA}{dt}$$

    When it is just about to move out:

    $$dA = 2R (vdt)$$

    $$ \cfrac{dA}{dt}= 2 Rv$$

    Hence, $$E = 2BRV$$
  • Question 8
    1 / -0
    Two coils have a mutual inductance $$0.005\ H$$. The current changes in the first coil according to the equation $$i = i_{m}\sin \omega t$$ where $$i_{m} = 10\ A$$ and $$\omega = 100\pi \ rad\ s^{-1}$$. The maximum value of the emf induced in the second coil is
    Solution
    Given :    $$L = 0.005 H$$   
    Variation of current in first coil    $$i = i_m \sin wt$$
    $$\therefore$$   $$\dfrac{di}{dt} = i_m w \cos wt$$
    Emf induced     $$\mathcal{E} = -L\dfrac{di}{dt} = -i_m w L \cos wt$$
    Thus maximum emf induced      $$\mathcal{E_{max}} = i_m wL$$             when  $$\cos wt  =-1$$
    $$\therefore$$     $$\mathcal{E_{max}} = 10\times 100\pi \times 0.005  = 5\pi$$  volts
  • Question 9
    1 / -0
    If emf induced in a coil is $$2V$$ by changing the current in it from $$8\ A$$ to $$6\ A$$ in $$2\times 10^{-3}s$$, then the coefficient of self induction is
    Solution
    Induced emf $$e = 2V$$

    $$i_{1} = 8A, i_{2} = 6A$$

    $$\triangle t = 2\times 10^{-3}s$$

    Coefficient of self induction

    $$L = \dfrac {e}{\triangle i/ \triangle t} = \dfrac {-2}{(6 - 8)/ 2\times 10^{-3}}$$

    $$= \dfrac {-2\times 2\times 10^{-3}}{-2}$$

    $$= 2\times 10^{-3} H$$
  • Question 10
    1 / -0
    A coil of resistance 5 unit and inductance 4 H is connected to a 10 V battery. The energy stored in thecoil
    Solution

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