Home
Class 12
PHYSICS
A circular coil of radius 10 cm, 500 tur...

A circular coil of radius 10 cm, 500 turns and resistance 2 Omega is placed with its plane prependicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through `180^(@)` in 0.25 s. Estimate the magnitude of the e.m.f and current induced in the coil. Horizotal component of earth's magnetic field at the place is `3 xx 10^(-5) T`.

Text Solution

Verified by Experts

Initial flux through the coil,
`Phi_(B(i n itial)=BAcos theta`
`=3.0xx10^(-5)xx(pixx10^(-2))xxcos0^(@)`
`=3pixx10^(-7)Wb`
Final flux after the rotation,
`Phi_(B(fi nal))=3.0xx10^(-5)xx(pixx10^(-2))xxcos180^(@)`
`= –3π × 10^(–7) Wb`
Therefore, estimated value of the induced emf is,
`epsi=N(DeltaPhi)/(Deltat)`
= 500xx(6pixx10^(–7))//0.25`
`=3.8xx10^(–3)V`
`I = epsi//R = 1.9xx10^(–3)`A
Note that the magnitudes of ε and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant.
Promotional Banner

Topper's Solved these Questions

  • ELECTROMAGNETIC INDUCTION

    NCERT|Exercise Exercise|17 Videos
  • ELECTRIC CHARGES AND FIELDS

    NCERT|Exercise EXERCISE|34 Videos
  • ELECTROMAGNETIC WAVES

    NCERT|Exercise Exercise|15 Videos

Similar Questions

Explore conceptually related problems

A circular coil having 500 turns and resistance 4 Omrga placed in a horizontal component of Earth's magnetic field of 4 xx 10^(-5) T induces an emf of 0.004 volt when rotated about its vertical diameter through 180^(@) in 0.15 s. Calculate the radius of the coil. Also find the induced current in the coil.

A circular coil of radius 20 cm and 500 turns and total resistance 5.0 Omega is placed with its plane vertical in such a manner that east-west line lies on its plane. Horizontal component of Earth's magnetic field is 30 mu tesla. Coil is rotated along its vertical diameter by an angle 180^(@) in 0.5 seconds. Calculate amount of charge circulated through the coil.

The horizontal component of earth's magnetic field is sqrt3 times the vertical component . What is the value of angle of dip at this place ?

The horizontal component of earth’s magnetic field at a place is sqrt""3 times the vertical component. The angle of dip at that place is

At a certain place , the horizontal component of earth's magnetic field is (1)/(sqrt3) times of its vertical component . The angle of dip at that place is

A coil of radius 10 cm and resistance 40 Omega has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180^@ . Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is B_H = 3.0X 10^(-5) T .

NCERT-ELECTROMAGNETIC INDUCTION-Exercise
  1. A circular coil of radius 10 cm, 500 turns and resistance 2 Omega is p...

    Text Solution

    |

  2. चित्र में वर्णित स्थितियों के लिए प्रेरित धारा की दिशा की प्रागुक्ति (...

    Text Solution

    |

  3. Use Lenz's law to determine the direction of induced current in the si...

    Text Solution

    |

  4. A long solenoid with 15 turns per cm has small loop of area 2.0 cm^(2)...

    Text Solution

    |

  5. A rectangular loop of sides 8 cm and 2 cm with a small cut is moving o...

    Text Solution

    |

  6. A 1m long calculating rod rotates with an angular frequency of 400 rad...

    Text Solution

    |

  7. A circular coil of radius 8.0 cm and 20 turns rotates about its vertic...

    Text Solution

    |

  8. A horizontal straight wire 10 m long extending from east to west falli...

    Text Solution

    |

  9. Current in a circuit falls form 0.5 A to 0.0 A in 0.1 s. If an average...

    Text Solution

    |

  10. A pair of adjacent coils has a mutual inductance of 1.5. H. If the cur...

    Text Solution

    |

  11. A jet plane is travelling west at the speed of 1800 km//h. What is the...

    Text Solution

    |

  12. Suppose the loop in above question is stationary, but the current feed...

    Text Solution

    |

  13. A square loop of side 12 cm with its sides parallel to X and Y axes is...

    Text Solution

    |

  14. It is desired to measure the magnitude of field between the poles of a...

    Text Solution

    |

  15. Fig. shows a metal rod PQ resting on the rails A, B and positoned betw...

    Text Solution

    |

  16. An air cored solenoid with length 30 cm, area of cross-section 25 cm^(...

    Text Solution

    |

  17. (a) Obtain an expression for the mutual inductance between a long stra...

    Text Solution

    |

  18. A line charge lambda per unit length is lodged uniformly onto the rim ...

    Text Solution

    |