Electromagnetic Induction

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Question
CBSEENPH12040053

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is

  • 1.0 Ω

  • 0.5 Ω

  • 2.0 Ω

  • zero

Solution

A.

1.0 Ω

This is a problem is a base on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential differences, then
E/V = l1/l2
where l1 and l2 are the length of potentiometer wire with and without short-circuited through a resistance

Since comma space
straight E over straight V space equals space fraction numerator straight R plus straight r over denominator straight R end fraction space left square bracket because space straight E space equals straight I space left parenthesis straight R plus straight r right parenthesis space and space straight V space equals IR right square bracket

therefore comma space fraction numerator straight R plus straight r over denominator straight R end fraction space equals straight l subscript 1 over straight l subscript 2
or space 1 space plus space straight r over straight R space equals space 110 over 100
straight r over straight R space equals space 10 over 100
straight r equals space 1 over 10 space straight x space 10 space equals space 1 straight capital omega

Sponsor Area

Question
CBSEENPH12040061

A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction in1 over straight pi open parentheses Wb over straight m squared close parentheses such a way that its axis makes an angle of 60o with.bold B with bold rightwards arrow on top The magnetic flux linked with the disc is 

  • 0.02 Wb

  • 0.06 Wb

  • 0.05 Wb

  • 0.01 Wb

Solution

A.

0.02 Wb

The magnetic flux  Φ passing through a plane surface of area A placed in a uniform magnetic field B is given by 
Φ = BA Cos θ
where θ is the angle between the direction of B and the normal to the plane.
Here,
straight theta space equals 60 to the power of straight o space comma space straight B space equals space 1 over straight pi space Wb divided by straight m squared comma space straight A space equals space straight pi space left parenthesis 0.2 right parenthesis squared
therefore comma space straight ϕ space equals space 1 over straight pi space straight x space straight pi space left parenthesis 0.2 right parenthesis squared space straight x space cos space 60
space equals space left parenthesis 0.2 right parenthesis squared space straight x space 1 half
space equals space 0.02 space Wb

Question
CBSEENPH12039840

A coil of resistance 400 is placed in magnetic field. If the magnetic flux Φ (Wb) linked with the coil varies with time t  (sec) as Φ = 50t2 +4.
The current in the coil at t= 2 s is

  • 0.5 A

  • 0.1 A

  • 2 A

  • 1 A

Solution

A.

0.5 A

 Induced space emf space of space space coil space straight E space equals space open vertical bar negative dΦ over dt close vertical bar subscript straight t
Given comma space equals space 50 straight t squared space plus 4 space and space straight R space equals space 400 straight capital omega
straight E equals space open vertical bar negative dΦ over dt close vertical bar subscript straight t space equals 2 end subscript space
equals open vertical bar 100 straight t close vertical bar subscript straight t equals 2 end subscript space equals space 200 straight V
Current space space in space the space coil space
straight i space equals space straight E over straight R space equals space 200 over 400
space equals space 0.5 space straight A space

Question
CBSEENPH12039901

A coil of self -inductance L is connected in series with a bulb b and an AC source. Brightness of the bulb decreases when

  • frequency of the AC source is decreases

  • number of turns in the coil is reduced

  • a capacitance of reactance Xc=XL is  included in the same circuit

  • an iron is inserted in the coil

Solution

D.

an iron is inserted in the coil

As space equals space straight Z space equals square root of straight R squared plus straight X subscript straight L superscript 2 end root space equals space square root of straight R squared plus left parenthesis 2 πfvL right parenthesis squared end root

As space straight I equals space straight V over straight Z. straight P space equals space straight I squared straight R
straight i. straight e. comma space straight V upwards arrow space straight L upwards arrow space rightwards double arrow space straight Z upwards arrow. space straight I downwards arrow space and space straight P downwards arrow

Question
CBSEENPH12040006

A condenser of capacity C is charged to a potential difference of V1. The plates to the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is

  • open parentheses fraction numerator straight C left parenthesis straight V subscript 1 minus straight V subscript 2 right parenthesis squared over denominator straight L end fraction close parentheses to the power of 1 divided by 2 end exponent
  • fraction numerator straight C left parenthesis straight V subscript 1 superscript 2 minus straight V subscript 2 superscript 2 right parenthesis over denominator straight L end fraction
  • fraction numerator straight C space left parenthesis straight V subscript 1 superscript 2 space plus straight V subscript 2 superscript 2 right parenthesis over denominator straight L end fraction
  • open parentheses fraction numerator straight C space left parenthesis straight V subscript 1 superscript 2 minus straight V subscript 2 superscript 2 right parenthesis over denominator straight L end fraction close parentheses to the power of 1 divided by 2 end exponent

Solution

D.

open parentheses fraction numerator straight C space left parenthesis straight V subscript 1 superscript 2 minus straight V subscript 2 superscript 2 right parenthesis over denominator straight L end fraction close parentheses to the power of 1 divided by 2 end exponent

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