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Electromagnetic Induction Test - 41

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Electromagnetic Induction Test - 41
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  • Question 1
    1 / -0
    A conducting rod AB moves parallel to X-axis in a uniform magnetic field, pointing in the positive X-direction. The end A of the rod gets

    Solution
    According to right hand palm rule, the Lorentz force on free electrons in the conductor will be directed towards and B. Hence, the end A gets positively charged.
  • Question 2
    1 / -0
    A coil in a simple generator has 200 turns.Now the number of turns in the coil increases to 500. What will be its effect:
    Solution
    By law of conservation of energy, mechanical power converted =electrical power produced assuming no losses 
    i.e. $$VI = $$ constant.
    When turns are increased, voltage induced increases, since voltage is proportional to to number of turns. 
    Hence, current reduces.
  • Question 3
    1 / -0
    Consider a current carrying coil placed in a magnetic field. What are the requirements for induced current to flow as per electromagnetic induction.
    Solution
    When magnetic flux through a coil changes, emf is induced in it. Current carrying coil is not required for above. 
  • Question 4
    1 / -0
    Which of the following will not increase voltage in a simple generator?
    Solution
    Voltage/current produced in a generator can be increased by:
    a) using a powerful electromagnet to make the magnetic field stronger in place of a permanent magnet.
    b) by winding the coil round a soft iron core to increase the strength of magnetic field.
    c) by using a coil with more turns.
    d) by rotating the coil faster.
    e) by using a coil with a larger area.
  • Question 5
    1 / -0
    A jet plane is travelling towards west at a speed of 1600 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earths magnetic field at the location has a magnitude of 5 $$\times 10^{-4}$$ T and the dip angle is $$30^o$$.
    Solution
    Vertical component of B,
    $$B \sin \delta = 5 \times 10^{-4} \times \sin (30)=2.5 \times 10^{-4} T$$
    EMF, $$E=Blv$$
    $$E=2.5 \times 10^{-4} \times 25 \times 1800 \times \dfrac{5}{18}=3.125 V$$
  • Question 6
    1 / -0
    The back emf in a DC motor is maximum when
    Solution
    The value of back emf is proportional to the motor speed. So, back emf is maximum when the motor has picked up the maximum speed.
  • Question 7
    1 / -0
    What is the self inductance of a coil in which an induced emf of $$2V$$ is set up, when the current is changing at the rate of $$4As^{-1}$$
    Solution
    Induced emf in a inductor $$L$$ is: $$\varepsilon = L \dfrac{di}{dt}= L  \times 4$$
    $$\Rightarrow 2 = L \times 4 $$
    $$\Rightarrow L = 0.5\: H$$
  • Question 8
    1 / -0
    You have a coil and a bar magnet. You can produce an electric current by:
    Solution
    Electric current can be produced by either moving a coil or a magnet and hence e.m.f is induced.
  • Question 9
    1 / -0
    The current produced in a closed coil, where magnetic lines of force rapidly change within it is called:
    Solution
    The property due to which a changing magnetic field within a closed conducting coil induces electric current in the coil is called electromagnetic induction. The current produced in a closed coil, where magnetic lines of force rapidly change within it is called Induced current. 
  • Question 10
    1 / -0
    A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is 15 m/s and is directed perpendicular to the field. The emf induced between the two ends of the conductor is:
    Solution
    Given, length of conductor l= 0.1 m
    Magnetic field  = $$0.1T$$
    Velocity = $$15m/s$$

    Emf Induced = $$Blv$$ = $$ 0.1X0.1X15 = 0.15V $$
    l=0.1 m
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