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Electromagnetic Induction Test - 42

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Electromagnetic Induction Test - 42
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  • Question 1
    1 / -0
    When the coil and the magnet are both stationary
    Solution
    When the coil and the magnet are both stationary, there is no deflection in the galvanometer. It is, thus, clear from this activity that motion of a magnet with respect to the coil produces an induced potential difference, which sets up an induced electric current in the circuit.
  • Question 2
    1 / -0
    The instrument which works on the principle of mutual inductance is
    Solution
    A Transformer works on the principle of mutual inductance.
    The mutual inductance between primary coil and secondary coil is responsible for transformation of the input voltage.
  • Question 3
    1 / -0
    For long distance transmission, the AC is stepped up because at high voltage, the transmission is
    Solution
    The transmission is undamped, because energy loss becomes is negligible.
  • Question 4
    1 / -0
    A machine is run such that electrical current is input and as a result rotation is achieved in the rotor. If current is turned off and rotation is forced in the machine
    Solution
    A motor behaves as a generator is input is reversed. Hence if current is removed and rotation is forced, electric current is induced in the coil due to electromagnetic induction. 
    By lenz's law, current is produced in a direction so as to oppose the changing flux through the coil. Hence, magnetic force on the coil is in a direction that opposes the motion of the coil and it experiences a resisting force.
  • Question 5
    1 / -0
    When current in a coil changes from $$+2 A$$ to $$-2 A$$ in $$0.05 s$$ and emf of $$8 V$$ is induced in the coil self-inductance of the coil is:
    Solution
    We know that the voltage across the inductor is given as:
    $$e= \dfrac{LdI}{dt}$$

    or $$L=\dfrac { edt }{ dI } $$

    Substituting the values in above expression:
    $$L=\dfrac { 8\times 0.05 }{ 2 } $$

    $$\Rightarrow L=0.2H$$
  • Question 6
    1 / -0
    The magnetic flux linked with a coil at any instant $$t$$ is given by the equation $$\phi =5{ t }^{ 3 }-100t+300$$. The magnitude of emf induced in the coil after $$3 s$$ is
    Solution
    Let us assume that the coil is made of a single turn only. 
    The emf induced is given by Faraday's law:
    $$\Rightarrow E=N\dfrac { d\phi  }{ dt } =15{ t }^{ 2 }-100$$

    Thus, after 3s, induced emf will be $$135-100=35V$$
  • Question 7
    1 / -0
    When a rod of length l is rotated with angular velocity of $$\omega$$ in a perpendicular field of induction B, about one end, the emf across its ends is
    Solution
    Consider an infinitesimally small length of rod at a distance $$x$$ from the center of the circular path.
    The emf induced across this element $$d\mathcal{E}=$$ $$Bv(dx)$$
    $$d\mathcal{E}=$$ $$B(x\omega)(dx) = Bw xdx$$
    Thus the net EMF induced $$\mathcal{E}$$=$$\int^l_0B\omega xdx$$
    $$\therefore$$  $$\mathcal{E}$$=$$ Bw  \int^l_0 xdx  = Bw \times \dfrac{x^2}{2}\bigg|^l_0$$
    $$\implies$$  $$\mathcal{E}$$=$$  Bw \times \dfrac{(l^2 - 0)}{2}$$ $$=\dfrac{Bl^2\omega}{2}$$

  • Question 8
    1 / -0
    The magnetic flux linked with a coil satisfies the relation $$\phi = 4t^{2} + 6t + 9$$ Wb, where t is the time in second. The emf induced in the coil at $$t = 2s$$ is
    Solution
    It is given that: $$\phi=4t^2+6t+9\ Wb$$
    Emf induced in the coil is equal to the rate of change of flux through it.
    Thus, emf induced $$=\dfrac{d\phi}{dt}=8t+6$$
    Thus, emf induced at $$(t=2s)=8(2)+6=22\ V$$
  • Question 9
    1 / -0
    The phenomenon of electromagnetic induction is :
    Solution
    Hint: When there is a change in the magnetic flux through a closed loop, an induced current is generated in such a way that it tries to oppose the change in the flux

    $$\textbf{Step 1: Discussing the electromagnetic induction}$$
    $$\bullet$$Electromagnetic induction is a phenomenon in which an emf is generated if there is a change in the magnetic flux through a closed-loop. 
    $$\bullet$$When there is relative motion between a magnet and the coil, magnetic flux changes, and hence an electromotive force is generated in the coil. This electromotive force generates induced current.

    $$\textbf{Therefore, option (C) is the correct answer.}$$
  • Question 10
    1 / -0
    What will be the self inductane of a coil of $$100$$ turns if a current of $$5\;A$$ produces a magnetic flux $$5\times10^{-5}$$ Wb?
    Solution
    GIven :    $$N = 100$$ turns              $$I = 5$$ $$A$$                 $$\phi  = 5 \times 10^{-5}$$ $$Wb$$
    Using                $$\phi_T  = LI$$            where  $$L$$ is  the self inductance
    $$\therefore$$   $$n \times \phi = LI$$
    $$100 \times 5 \times 10^{-5}  = L \times 5$$              $$\implies  L =10^{-3} H  = 1mH$$
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