Electromagnetic Induction

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Question
CBSEENPH12039436

State the principle of an ac generators and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil havin N turns each of cross-sectional area, rotating with a constant angular speed 'ω' in a magnetic field B, directed perpendicular to the axis of rotation.

Solution

Principle − AC generator based on the phenomenon of electromagnetic induction.

Construction:

Main parts of an ac generator:

  1. Armature − Rectangular coil ABCD

  2. Filed Magnets − Two pole pieces of a strong electromagnet

  3. Slip Rings − The ends of coil ABCD are connected to two hollow metallic rings R1 and R2.

  4. Brushes − B1 and B2 are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R1 and R2 respectively.

Theory and Working − As the armature coil is rotated in the magnetic field, angle θ between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An emf is induced in the coil. According to Fleming’s right-hand rule, current induced in AB is from A to B and it is from C to D in CD. In the external circuit, current flows from B2 to B1.

To calculate the magnitude of emf induced:

Suppose

A → Area of each turn of the coil

N → Number of turns in the coil

B→ Strength of magnetic field

θ → Angle which normal to the coil makes with State the principle of an ac generators and explain its working with tat any instant t

State the principle of an ac generators and explain its working with t

∴ Magnetic flux linked with the coil in this position:

Φ = N (B.A)= NBA cosθ= NBA cosωt …(i)

Where, ‘ω’ is angular velocity of the coil

As the coil rotates, angle θchanges. Therefore, magnetic flux Φ linked with the coil changes and hence, an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then

e = - dt =-ddt (NAB cos ωt) = -NABddt(cos ωt)=-NAB (-sin ωt) ω

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Question
CBSEENPH12039437

An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10–4 T and the angle of dip is 30°.

Solution

e = Bv.V.l

=5 x 10-4 sin 30 x 900 x 518 x 20 = 1.25 V

Question
CBSEENPH12039596

An inductor of inductance L = 400 mH and resistors of resistances R1 = 2 Ω and R2 = 2 Ω is connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is 

  • 6e–5t V

  • 12 over straight t straight e to the power of negative 3 straight t end exponent
  • 6 space open parentheses 1 minus straight e to the power of fraction numerator negative straight t over denominator 0.2 end fraction end exponent close parentheses space straight V
  • 12e–5t V

Solution

D.

12e–5t V

straight e subscript straight L space equals space Ee to the power of negative tR subscript 2 over straight L end exponent
straight e subscript straight L space equals space 12 straight e to the power of negative 2 over 4 straight t end exponent
straight e subscript straight L space equals space 12 straight e to the power of negative 5 straight t end exponent

Question
CBSEENPH12039633

Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I1 and COD carries a current I2. The magnetic field on a point lying at a distance ‘d’ from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by

  • fraction numerator straight mu subscript 0 over denominator 2 straight pi end fraction space open parentheses fraction numerator straight I subscript 1 plus straight I subscript 2 over denominator straight d end fraction close parentheses to the power of 1 divided by 2 end exponent
  • fraction numerator straight mu subscript 0 over denominator 2 πd end fraction space left parenthesis straight l subscript 1 superscript 2 space plus space straight I subscript 2 superscript 2 right parenthesis to the power of 1 divided by 2 end exponent
  • fraction numerator straight mu subscript 0 over denominator 2 πd end fraction left parenthesis straight I subscript 1 space plus straight I subscript 2 right parenthesis
  • fraction numerator straight mu subscript 0 over denominator 2 πd end fraction left parenthesis straight I subscript 1 superscript 2 space plus space straight I subscript 2 superscript 2 right parenthesis

Solution

B.

fraction numerator straight mu subscript 0 over denominator 2 πd end fraction space left parenthesis straight l subscript 1 superscript 2 space plus space straight I subscript 2 superscript 2 right parenthesis to the power of 1 divided by 2 end exponent