Home
Class 12
PHYSICS
Find the velocity of charge leading to 1...

Find the velocity of charge leading to `1A` current which flows in a copper conductor of cross section `1cm^2` and length `10 km`. Free electron density of copper is `8.5x10^28//m^3`. How long will it take the electric charge to travel from one end of the conductor to the other?

Promotional Banner

Topper's Solved these Questions

  • CAPACITANCE

    PHYSICS GALAXY - ASHISH ARORA|Exercise UNSOLVED NUMERICAL PROBLEMS|40 Videos
  • ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT

    PHYSICS GALAXY - ASHISH ARORA|Exercise Advance MCQs|33 Videos

Similar Questions

Explore conceptually related problems

Calculate the drift speed of the electrons when 1A of current exists in a copper wire of cross section 2mm^2 The number of free electrons in 1cm^3 of copper is 8.5xx10^22

A current of 10 A is maintained in a conductor of cross-section 1 cm^(2) . If the free electron density in the conductor is 9 xx 10^(28) m^(-3) , then drift velocity of free electrons is

Calculate the drift speed of the electrons when 1A of current exists in a copper wire of cross section 2 mm^(2) .The number of free electrons in 1cm^(3) of copper is 8.5xx10^(22) .

A current 0.5 amperes flows in a conductor of cross-sectional area of 10^(-2) m^2 . If the electron density is 0.3 * 10^(28) m^(-3) , then the drift velocity of free electrons is

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross setion mm^(2) , is v. If the electron density in copper is 9xx10^(28)//m^(3) the value of v in mm//s close to (Take charge of electron to be =1.6xx10^(-19)C)

A current of 4.8 A is flowing in a copper wire of cross-sectional area 3xx10^(-4) m^(2) . Find the current density in the wire.

Find the average drift speed of free electrons in a copper wire of area of cross-section 10^(-7) m^(2) carrying current of 1.5 A and having free electron density 8.5 xx 10^(28) m^(-3)

PHYSICS GALAXY - ASHISH ARORA-CURRENT ELECTRICITY-All Questions
  1. Find the velocity of charge leading to 1A current which flows in a co...

    Text Solution

    |

  2. The potential difference across a straight wire of 10^(-3) cm^(2) cros...

    Text Solution

    |

  3. Find the total linear momentum of the electrons in a conductor of leng...

    Text Solution

    |

  4. Two large parallel metal plartes are located in vacuum. One of these s...

    Text Solution

    |

  5. A copper wire has a square cross section of 6 mm on a side. The wire i...

    Text Solution

    |

  6. How many electrons per second pass through a section of resistance of ...

    Text Solution

    |

  7. 1 meter long metallic wire is broken into two unequal parts P and Q. ...

    Text Solution

    |

  8. The area of cross-section, length and density of a piece of a metal of...

    Text Solution

    |

  9. Find equivalent resistant across terminals A and B in the circuit show...

    Text Solution

    |

  10. Find equivalent resistance of the circuit shown in figure 3.32 across ...

    Text Solution

    |

  11. Calculate battery current and equivalent resistance of the network sho...

    Text Solution

    |

  12. A long resistor between point A and B as shown in figure-3.35 has reis...

    Text Solution

    |

  13. You need to produce a set of cylindrical copper wires 2.5 m long that ...

    Text Solution

    |

  14. Two resistors with temperature coefficients of resistance alpha1 and a...

    Text Solution

    |

  15. Two coils connected in series have resistance of 600K Omega and 300 Om...

    Text Solution

    |

  16. Shown a conductor of length l having a circular cross section. The rad...

    Text Solution

    |

  17. ABCD is a square where each side is a uniform wire of resistance 1Omeg...

    Text Solution

    |

  18. Four resistors R(1),R(2),R(3) and R(4) are connected between two termi...

    Text Solution

    |

  19. In fig the steady state current in 2omega resistance is

    Text Solution

    |

  20. Find the equivalent resistance and current in 6Omega resistance in the...

    Text Solution

    |