Electromagnetic Induction

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Question 1

A boat is moving due east in a region where the earth's magnetic field is 5.0 × 10-5 NA-1m-1 due north and horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is 1.50 ms-1, the magnitude of the induced emf in the wire of aerial is

  • 0.75 mV

  • 0.50 mV

  • 0.15 mV

  • 1 mV

Solution

C.

0.15 mV

Eind = B × v × l
= 5.0 × 10-5 × 1.50 × 2
= 10.0 × 10-5 × 1.5
= 15 × 10-5 vot.
= 0.15 mv

Question 2

A coil having n turns and resistance 4R Ω. This combination is moved in time t seconds from a magnetic field W1 weber to W2 weber. The induced current in the circuit is

  • fraction numerator negative straight W subscript 2 minus straight W subscript 1 over denominator 5 Rnt end fraction
  • fraction numerator negative left parenthesis straight W subscript 2 minus straight W subscript 1 right parenthesis over denominator 5 Rt end fraction
  • fraction numerator negative straight W subscript 2 minus straight W subscript 1 over denominator Rnt end fraction
  • fraction numerator straight n left parenthesis straight W subscript 2 minus straight W subscript 1 right parenthesis over denominator Rt end fraction

Solution

B.

fraction numerator negative left parenthesis straight W subscript 2 minus straight W subscript 1 right parenthesis over denominator 5 Rt end fraction straight I equals space straight n over straight R comma dϕ over dt
or space straight I equals space 1 over straight R comma space straight n space open square brackets fraction numerator straight W subscript 2 minus straight W subscript 1 over denominator straight t subscript 2 minus space straight t subscript 1 end fraction close square brackets
straight I equals space fraction numerator negative 1 over denominator left parenthesis straight R space plus 4 straight R right parenthesis end fraction fraction numerator straight n left parenthesis straight W subscript 2 minus straight W subscript 1 right parenthesis over denominator straight t end fraction
or comma space straight I space equals space fraction numerator straight n left parenthesis straight W subscript 2 minus straight W subscript 1 right parenthesis over denominator 5 Rt end fraction
(W1 and W2 are not the magnetic fields, but the values of flux associated with one turn of coil)
Question 3

A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to

  • development of air current when the plate is placed.

  • induction of electrical charge on the plate

  • shielding of magnetic lines of force as aluminium is a paramagnetic material.

  • electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

Solution

D.

electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

The oscillating coil produces time variable magnetic field. It causes eddy current in the aluminium plate which causes anti–torque on the coil, due to which it stops.

Question 4

A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage 2V. The current reaches half of its steady state value in

  • 0.05 s

  • 0.1 s

  • 0.15 s

  • 0.3 s

Solution

B.

0.1 s

straight I space equals space straight I subscript straight o space open parentheses 1 minus straight e to the power of negative straight R over straight L straight t end exponent close parentheses
0.693 space equals straight R over straight L straight t
straight t space equals space fraction numerator 0.3 space straight x space.693 over denominator 2 end fraction space equals space 0.1 space sec
Question 5

A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth’s magnetic field is 0.3 × 10−4 T, then the e.m.f. developed between the two ends of the conductor is

  • depends on the nature of the metal used

  • depends on the intensity of the radiation

  • depends both on the intensity of the radiation and the metal used

  • is the same for all metals and independent of the intensity of the radiation.

Solution

B.

depends on the intensity of the radiation

The emf induced between ends of conductor
straight e space equals space 1 half BωL squared
space equals space 1 half space straight x space 0.2 space straight x space 10 to the power of negative 4 end exponent space straight x space 5 space straight x space left parenthesis 1 right parenthesis squared
space equals space 0.5 space straight x space 10 to the power of negative 4 end exponent space straight V
space equals space 5 space straight x space 10 to the power of negative 5 end exponent space straight V space equals space 50 space μV